{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2026-04-16T00:12:35.000Z","description":"Problems submitted by members of the MATLAB Central community.","is_default":true,"created_by":161519,"badge_id":null,"featured":false,"trending":false,"solution_count_in_trending_period":0,"trending_last_calculated":"2026-04-16T00:00:00.000Z","image_id":null,"published":true,"community_created":false,"status_id":2,"is_default_group_for_player":false,"deleted_by":null,"deleted_at":null,"restored_by":null,"restored_at":null,"description_opc":null,"description_html":null,"published_at":null},"problems":[{"id":45184,"title":"factorial","description":"calculate x!","description_html":"\u003cp\u003ecalculate x!\u003c/p\u003e","function_template":"function y = your_fcn_name(x)\r\ny=fact(x)\r\nend","test_suite":"%%\r\nx = 1;\r\ny_correct = 1;\r\nassert(isequal(your_fcn_name(x),y_correct))\r\n%%\r\nx = 2;\r\ny_correct = 2;\r\nassert(isequal(your_fcn_name(x),y_correct))\r\n%%\r\nx = 3;\r\ny_correct = 6;\r\nassert(isequal(your_fcn_name(x),y_correct))\r\n%%\r\nx = 4;\r\ny_correct = 24;\r\nassert(isequal(your_fcn_name(x),y_correct))\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":368653,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":71,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2019-10-23T10:14:44.000Z","updated_at":"2026-02-18T21:42:08.000Z","published_at":"2019-10-23T10:14:46.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ecalculate x!\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":44871,"title":"Calculate Resistance","description":"Calculate Resistance R of a linear conductor having resistivity p, length l and area A","description_html":"\u003cp\u003eCalculate Resistance R of a linear conductor having resistivity p, length l and area A\u003c/p\u003e","function_template":"function R = Resistance(p,l,A);\r\nR=*p+l/A^2;\r\n","test_suite":"%%\r\np=1;l=1;A=2;\r\ncorrect = 0.5;\r\nassert(isequal(Resistance(p,l,A),correct))\r\n%%\r\np=0.5;l=10;A=2;\r\ncorrect = 2.5;\r\nassert(isequal(Resistance (p,l,A),correct))","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":293792,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":105,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2019-03-07T08:28:08.000Z","updated_at":"2026-03-03T22:21:36.000Z","published_at":"2019-03-07T08:28:08.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eCalculate Resistance R of a linear conductor having resistivity p, length l and area A\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":44872,"title":"Calculate Resistance 2","description":"In this problem, you have to calculate Resistance R of a linear conductor  having voltage V across it and current I is passing in it.","description_html":"\u003cp\u003eIn this problem, you have to calculate Resistance R of a linear conductor  having voltage V across it and current I is passing in it.\u003c/p\u003e","function_template":"function y = Resistance(V,I)\r\n  y = x;\r\nend","test_suite":"%%\r\nV=4;I=8;\r\ny_correct = 0.5;\r\nassert(isequal(Resistance(V,I),y_correct))\r\n%%\r\nV=7;I=1.75;\r\ny_correct = 4;\r\nassert(isequal(Resistance(V,I),y_correct))","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":293792,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":97,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2019-03-07T08:43:11.000Z","updated_at":"2026-02-09T15:00:57.000Z","published_at":"2019-03-07T08:43:11.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eIn this problem, you have to calculate Resistance R of a linear conductor having voltage V across it and current I is passing in it.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":44865,"title":"Efficiency of a single phase Transformer","description":"Calculate the efficiency of a single phase transformer whose KVA rating is A KVA, loading factor x,power factor p,full load copper loss Pcu Kilowatt, iron loss Pi Kilowatt. Efficiency = output power/input power","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.4333px; min-height: 0px; white-space: normal; color: rgb(0, 0, 0); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: rgb(0, 0, 0); white-space: normal; \"\u003e\u003cdiv style=\"block-size: 42px; display: block; min-width: 0px; padding-block-start: 0px; padding-top: 0px; perspective-origin: 407px 21px; transform-origin: 407px 21px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; perspective-origin: 384px 21px; text-align: left; transform-origin: 384px 21px; white-space: pre-wrap; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 380.5px 8px; transform-origin: 380.5px 8px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eCalculate the efficiency of a single phase transformer whose KVA rating is A KVA, loading factor x,power factor p,full load copper loss Pcu Kilowatt, iron loss Pi Kilowatt. Efficiency = output power/input power\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function y = efficiency(A,x,p,Pcu,Pi)\r\n  y = A/(A+25*x);\r\nend","test_suite":"%%\r\nA=5; x=0.5; p=0.5; Pcu=0; Pi=0;\r\ny_correct = 1;\r\nassert(isequal(efficiency(A,x,p,Pcu,Pi),y_correct))\r\n\r\n%%\r\nA=1; x=1; p=1; Pcu=1; Pi=0;\r\ny_correct = 0.5;\r\nassert(isequal(efficiency(A,x,p,Pcu,Pi),y_correct))\r\n\r\n%%\r\nA=randi(10); x=randi(10); p=0; Pcu=randi(10); Pi=randi(10);\r\ny_correct = 0;\r\nassert(isequal(efficiency(A,x,p,Pcu,Pi),y_correct))","published":true,"deleted":false,"likes_count":1,"comments_count":2,"created_by":293792,"edited_by":223089,"edited_at":"2022-10-27T13:31:33.000Z","deleted_by":null,"deleted_at":null,"solvers_count":19,"test_suite_updated_at":"2022-10-27T13:31:33.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2019-03-07T06:31:45.000Z","updated_at":"2026-03-11T15:36:42.000Z","published_at":"2019-03-07T06:31:45.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eCalculate the efficiency of a single phase transformer whose KVA rating is A KVA, loading factor x,power factor p,full load copper loss Pcu Kilowatt, iron loss Pi Kilowatt. Efficiency = output power/input power\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":496,"title":"Oxidation State","description":"* In some chemical compounds, the oxidation state of each atom of hydrogen H is +1.\r\n* In some chemical compounds, the oxidation state of each atom of oxygen O is -2.\r\n* The algebraic sum of oxidation states of all atoms in a neutral molecule must be zero.\r\n* Given a string such as 'sulphuric acid H2SO4', the final token represents the chemical formula of a neutral moledule. This molecule has two atoms of hydrogen, one atom of sulphur and four atoms of oxygen. In this case the oxidation state of sulphur is six, 0-(2*1+4*-2). \r\n* Another string may be 'sodium hydroxide NaOH'. This molecule has one atom of sodium, one atom of oxygen, and, one atom of hydrogen. In this case, the oxidation state of sodium is one, 0-(-2+1). \r\n* Each element is denoted by one upper case letter, followed by one or none lower case letter, followed by the number of atoms present, no number means one atom.  \r\n* Assuming above, please find the oxidation state of one atom of the new element in the given formula.\r\n* Here is a link for additional information and exceptions to these simple rules: \u003chttp://en.wikipedia.org/wiki/Oxidation_state Oxidation State\u003e. \r\n","description_html":"\u003cul\u003e\u003cli\u003eIn some chemical compounds, the oxidation state of each atom of hydrogen H is +1.\u003c/li\u003e\u003cli\u003eIn some chemical compounds, the oxidation state of each atom of oxygen O is -2.\u003c/li\u003e\u003cli\u003eThe algebraic sum of oxidation states of all atoms in a neutral molecule must be zero.\u003c/li\u003e\u003cli\u003eGiven a string such as 'sulphuric acid H2SO4', the final token represents the chemical formula of a neutral moledule. This molecule has two atoms of hydrogen, one atom of sulphur and four atoms of oxygen. In this case the oxidation state of sulphur is six, 0-(2*1+4*-2).\u003c/li\u003e\u003cli\u003eAnother string may be 'sodium hydroxide NaOH'. This molecule has one atom of sodium, one atom of oxygen, and, one atom of hydrogen. In this case, the oxidation state of sodium is one, 0-(-2+1).\u003c/li\u003e\u003cli\u003eEach element is denoted by one upper case letter, followed by one or none lower case letter, followed by the number of atoms present, no number means one atom.\u003c/li\u003e\u003cli\u003eAssuming above, please find the oxidation state of one atom of the new element in the given formula.\u003c/li\u003e\u003cli\u003eHere is a link for additional information and exceptions to these simple rules: \u003ca href=\"http://en.wikipedia.org/wiki/Oxidation_state\"\u003eOxidation State\u003c/a\u003e.\u003c/li\u003e\u003c/ul\u003e","function_template":"function y = oxno(x)\r\nx='sulphuric acid H2SO4';\r\ny = 0-(2*1+4*-2);\r\nend","test_suite":"%%\r\nx='sulphuric acid H2SO4';\r\ny = 0-(2*1+4*-2);\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='sodium hydroxide NaOH';\r\ny = 0-(-2+1);\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='methane CH4';\r\ny = 0-(4*1);\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='ethane C2H6';\r\ny = 0-(6*1)/2;\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='acetylene C2H2';\r\ny = 0-(2*1)/2;\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='hypophosphorous acid H3PO2';\r\ny = 1;\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='(ortho)phosphorous acid H3PO3';\r\ny = 3;\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='pyrophosphoric acid H4P2O7';\r\ny = 5;\r\nassert(isequal(y,oxno(x)))\r\n\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":166,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":20,"test_suite_updated_at":"2012-03-14T02:03:22.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2012-03-14T00:49:41.000Z","updated_at":"2026-01-22T12:38:01.000Z","published_at":"2012-03-14T03:11:26.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eIn some chemical compounds, the oxidation state of each atom of hydrogen H is +1.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eIn some chemical compounds, the oxidation state of each atom of oxygen O is -2.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe algebraic sum of oxidation states of all atoms in a neutral molecule must be zero.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven a string such as 'sulphuric acid H2SO4', the final token represents the chemical formula of a neutral moledule. This molecule has two atoms of hydrogen, one atom of sulphur and four atoms of oxygen. In this case the oxidation state of sulphur is six, 0-(2*1+4*-2).\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eAnother string may be 'sodium hydroxide NaOH'. This molecule has one atom of sodium, one atom of oxygen, and, one atom of hydrogen. In this case, the oxidation state of sodium is one, 0-(-2+1).\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eEach element is denoted by one upper case letter, followed by one or none lower case letter, followed by the number of atoms present, no number means one atom.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eAssuming above, please find the oxidation state of one atom of the new element in the given formula.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eHere is a link for additional information and exceptions to these simple rules:\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:hyperlink w:docLocation=\\\"http://en.wikipedia.org/wiki/Oxidation_state\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eOxidation State\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"}],"problem_search":{"errors":[],"problems":[{"id":45184,"title":"factorial","description":"calculate x!","description_html":"\u003cp\u003ecalculate x!\u003c/p\u003e","function_template":"function y = your_fcn_name(x)\r\ny=fact(x)\r\nend","test_suite":"%%\r\nx = 1;\r\ny_correct = 1;\r\nassert(isequal(your_fcn_name(x),y_correct))\r\n%%\r\nx = 2;\r\ny_correct = 2;\r\nassert(isequal(your_fcn_name(x),y_correct))\r\n%%\r\nx = 3;\r\ny_correct = 6;\r\nassert(isequal(your_fcn_name(x),y_correct))\r\n%%\r\nx = 4;\r\ny_correct = 24;\r\nassert(isequal(your_fcn_name(x),y_correct))\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":368653,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":71,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2019-10-23T10:14:44.000Z","updated_at":"2026-02-18T21:42:08.000Z","published_at":"2019-10-23T10:14:46.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ecalculate x!\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":44871,"title":"Calculate Resistance","description":"Calculate Resistance R of a linear conductor having resistivity p, length l and area A","description_html":"\u003cp\u003eCalculate Resistance R of a linear conductor having resistivity p, length l and area A\u003c/p\u003e","function_template":"function R = Resistance(p,l,A);\r\nR=*p+l/A^2;\r\n","test_suite":"%%\r\np=1;l=1;A=2;\r\ncorrect = 0.5;\r\nassert(isequal(Resistance(p,l,A),correct))\r\n%%\r\np=0.5;l=10;A=2;\r\ncorrect = 2.5;\r\nassert(isequal(Resistance (p,l,A),correct))","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":293792,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":105,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2019-03-07T08:28:08.000Z","updated_at":"2026-03-03T22:21:36.000Z","published_at":"2019-03-07T08:28:08.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eCalculate Resistance R of a linear conductor having resistivity p, length l and area A\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":44872,"title":"Calculate Resistance 2","description":"In this problem, you have to calculate Resistance R of a linear conductor  having voltage V across it and current I is passing in it.","description_html":"\u003cp\u003eIn this problem, you have to calculate Resistance R of a linear conductor  having voltage V across it and current I is passing in it.\u003c/p\u003e","function_template":"function y = Resistance(V,I)\r\n  y = x;\r\nend","test_suite":"%%\r\nV=4;I=8;\r\ny_correct = 0.5;\r\nassert(isequal(Resistance(V,I),y_correct))\r\n%%\r\nV=7;I=1.75;\r\ny_correct = 4;\r\nassert(isequal(Resistance(V,I),y_correct))","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":293792,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":97,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2019-03-07T08:43:11.000Z","updated_at":"2026-02-09T15:00:57.000Z","published_at":"2019-03-07T08:43:11.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eIn this problem, you have to calculate Resistance R of a linear conductor having voltage V across it and current I is passing in it.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":44865,"title":"Efficiency of a single phase Transformer","description":"Calculate the efficiency of a single phase transformer whose KVA rating is A KVA, loading factor x,power factor p,full load copper loss Pcu Kilowatt, iron loss Pi Kilowatt. Efficiency = output power/input power","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.4333px; min-height: 0px; white-space: normal; color: rgb(0, 0, 0); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: rgb(0, 0, 0); white-space: normal; \"\u003e\u003cdiv style=\"block-size: 42px; display: block; min-width: 0px; padding-block-start: 0px; padding-top: 0px; perspective-origin: 407px 21px; transform-origin: 407px 21px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; perspective-origin: 384px 21px; text-align: left; transform-origin: 384px 21px; white-space: pre-wrap; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 380.5px 8px; transform-origin: 380.5px 8px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eCalculate the efficiency of a single phase transformer whose KVA rating is A KVA, loading factor x,power factor p,full load copper loss Pcu Kilowatt, iron loss Pi Kilowatt. Efficiency = output power/input power\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function y = efficiency(A,x,p,Pcu,Pi)\r\n  y = A/(A+25*x);\r\nend","test_suite":"%%\r\nA=5; x=0.5; p=0.5; Pcu=0; Pi=0;\r\ny_correct = 1;\r\nassert(isequal(efficiency(A,x,p,Pcu,Pi),y_correct))\r\n\r\n%%\r\nA=1; x=1; p=1; Pcu=1; Pi=0;\r\ny_correct = 0.5;\r\nassert(isequal(efficiency(A,x,p,Pcu,Pi),y_correct))\r\n\r\n%%\r\nA=randi(10); x=randi(10); p=0; Pcu=randi(10); Pi=randi(10);\r\ny_correct = 0;\r\nassert(isequal(efficiency(A,x,p,Pcu,Pi),y_correct))","published":true,"deleted":false,"likes_count":1,"comments_count":2,"created_by":293792,"edited_by":223089,"edited_at":"2022-10-27T13:31:33.000Z","deleted_by":null,"deleted_at":null,"solvers_count":19,"test_suite_updated_at":"2022-10-27T13:31:33.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2019-03-07T06:31:45.000Z","updated_at":"2026-03-11T15:36:42.000Z","published_at":"2019-03-07T06:31:45.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eCalculate the efficiency of a single phase transformer whose KVA rating is A KVA, loading factor x,power factor p,full load copper loss Pcu Kilowatt, iron loss Pi Kilowatt. Efficiency = output power/input power\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":496,"title":"Oxidation State","description":"* In some chemical compounds, the oxidation state of each atom of hydrogen H is +1.\r\n* In some chemical compounds, the oxidation state of each atom of oxygen O is -2.\r\n* The algebraic sum of oxidation states of all atoms in a neutral molecule must be zero.\r\n* Given a string such as 'sulphuric acid H2SO4', the final token represents the chemical formula of a neutral moledule. This molecule has two atoms of hydrogen, one atom of sulphur and four atoms of oxygen. In this case the oxidation state of sulphur is six, 0-(2*1+4*-2). \r\n* Another string may be 'sodium hydroxide NaOH'. This molecule has one atom of sodium, one atom of oxygen, and, one atom of hydrogen. In this case, the oxidation state of sodium is one, 0-(-2+1). \r\n* Each element is denoted by one upper case letter, followed by one or none lower case letter, followed by the number of atoms present, no number means one atom.  \r\n* Assuming above, please find the oxidation state of one atom of the new element in the given formula.\r\n* Here is a link for additional information and exceptions to these simple rules: \u003chttp://en.wikipedia.org/wiki/Oxidation_state Oxidation State\u003e. \r\n","description_html":"\u003cul\u003e\u003cli\u003eIn some chemical compounds, the oxidation state of each atom of hydrogen H is +1.\u003c/li\u003e\u003cli\u003eIn some chemical compounds, the oxidation state of each atom of oxygen O is -2.\u003c/li\u003e\u003cli\u003eThe algebraic sum of oxidation states of all atoms in a neutral molecule must be zero.\u003c/li\u003e\u003cli\u003eGiven a string such as 'sulphuric acid H2SO4', the final token represents the chemical formula of a neutral moledule. This molecule has two atoms of hydrogen, one atom of sulphur and four atoms of oxygen. In this case the oxidation state of sulphur is six, 0-(2*1+4*-2).\u003c/li\u003e\u003cli\u003eAnother string may be 'sodium hydroxide NaOH'. This molecule has one atom of sodium, one atom of oxygen, and, one atom of hydrogen. In this case, the oxidation state of sodium is one, 0-(-2+1).\u003c/li\u003e\u003cli\u003eEach element is denoted by one upper case letter, followed by one or none lower case letter, followed by the number of atoms present, no number means one atom.\u003c/li\u003e\u003cli\u003eAssuming above, please find the oxidation state of one atom of the new element in the given formula.\u003c/li\u003e\u003cli\u003eHere is a link for additional information and exceptions to these simple rules: \u003ca href=\"http://en.wikipedia.org/wiki/Oxidation_state\"\u003eOxidation State\u003c/a\u003e.\u003c/li\u003e\u003c/ul\u003e","function_template":"function y = oxno(x)\r\nx='sulphuric acid H2SO4';\r\ny = 0-(2*1+4*-2);\r\nend","test_suite":"%%\r\nx='sulphuric acid H2SO4';\r\ny = 0-(2*1+4*-2);\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='sodium hydroxide NaOH';\r\ny = 0-(-2+1);\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='methane CH4';\r\ny = 0-(4*1);\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='ethane C2H6';\r\ny = 0-(6*1)/2;\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='acetylene C2H2';\r\ny = 0-(2*1)/2;\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='hypophosphorous acid H3PO2';\r\ny = 1;\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='(ortho)phosphorous acid H3PO3';\r\ny = 3;\r\nassert(isequal(y,oxno(x)))\r\n%%\r\nx='pyrophosphoric acid H4P2O7';\r\ny = 5;\r\nassert(isequal(y,oxno(x)))\r\n\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":166,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":20,"test_suite_updated_at":"2012-03-14T02:03:22.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2012-03-14T00:49:41.000Z","updated_at":"2026-01-22T12:38:01.000Z","published_at":"2012-03-14T03:11:26.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eIn some chemical compounds, the oxidation state of each atom of hydrogen H is +1.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eIn some chemical compounds, the oxidation state of each atom of oxygen O is -2.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe algebraic sum of oxidation states of all atoms in a neutral molecule must be zero.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven a string such as 'sulphuric acid H2SO4', the final token represents the chemical formula of a neutral moledule. This molecule has two atoms of hydrogen, one atom of sulphur and four atoms of oxygen. In this case the oxidation state of sulphur is six, 0-(2*1+4*-2).\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eAnother string may be 'sodium hydroxide NaOH'. This molecule has one atom of sodium, one atom of oxygen, and, one atom of hydrogen. In this case, the oxidation state of sodium is one, 0-(-2+1).\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eEach element is denoted by one upper case letter, followed by one or none lower case letter, followed by the number of atoms present, no number means one atom.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eAssuming above, please find the oxidation state of one atom of the new element in the given formula.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eHere is a link for additional information and exceptions to these simple rules:\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:hyperlink w:docLocation=\\\"http://en.wikipedia.org/wiki/Oxidation_state\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eOxidation State\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray 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