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{"ops":[{"insert":"Is my answer right for this question?\n\nAssuming the conditions are at SATP, calculate both the theoretical volume and mass (in g) of both water and carbon dioxide produced (assuming all of the sodium bicarbonate decomposed).\u00a0 Use proper sig digs. Show all of your work.\u00a0 (\/4)"},{"attributes":{"list":"ordered"},"insert":"\n"},{"insert":"2 NaHCO3\u00a0 \u2192 \u00a0 Na2CO3\u00a0 + H2O\u00a0 + \u00a0 CO2\n\n\n3.95g Na2CO3 x 1 mol Na2CO3 \/ 84.01g Na2CO3 = 0.0470 mol Na2CO3\n\n"},{"attributes":{"italic":true,"bold":true},"insert":"H2O"},{"insert":"\n0.0470 mol Na2CO3 x 1 mol H2O \/ 1 mol Na2CO3 = 0.0470 mol H2O\n0.0470 mol H2O x 18.02 g H2O \/ 1 mol H2O = "},{"attributes":{"bold":true},"insert":"0.847g H2O (MASS, 3SD)"},{"insert":"\n"},{"attributes":{"bold":true},"insert":"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(1.01 x 2 + 16.00)"},{"insert":"\n0.847g H2O x 1ml H2O \/ 1g H2O = "},{"attributes":{"italic":true,"bold":true},"insert":"0.847ml H2O (VOLUME, 4SD)"},{"insert":"\n\n"},{"attributes":{"italic":true,"bold":true},"insert":"CO2"},{"insert":"\n0.0470 mol Na2CO3 x 1 mol CO2 \/ 1 mol Na2CO3 = 0.0470 mol CO2 "},{"attributes":{"bold":true},"insert":"(n, gas law)"},{"insert":"\n0.0470 mol CO2 x 44.01g CO2 \/ 1 mol CO2 = "},{"attributes":{"italic":true,"bold":true},"insert":"2.07g CO2 (MASS, 3SD)"},{"insert":"\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(16.00 x 2 + 12.01)\nPV = nRT\u00a0\nV = nRT \/ P\u00a0\nV = (0.0470 mol CO2) (8.314) (24.8) \/ 101.325\u00a0\n"},{"attributes":{"bold":true},"insert":"V = 0.0956L CO2 "},{"attributes":{"italic":true,"bold":true},"insert":"(VOLUME, 3SD)"},{"attributes":{"bold":true},"insert":"\u00a0"},{"insert":"\n"}]}

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