write a balanced equation for the combustion of acetaldehyde, CH3CHO?
962190
C2H4O + 5/2O2 ----> 2CO2 + 2H2O, or get rid of the fractions to get:\n2C2H4O + 5O2 ----> 4CO2 + 4H2O
868179
2Rb + 2 H20 --> 2RbOH + H2
851681
Write the balanced equation - you need the mole ratios.\n\nChange 1g hexane to moles of hexane.\n\nSee how many moles of oxygen are USED and how many moles of  CO2 and H2O are formed.\n\nYour total number of moles in the reaction mixture would then be:\n\nmoles of Oxygen NOT used + moles of CO2 (we will assume the water formed consenses to a liquid at 20 degrees).\n\nUse the Ideal Gas Law to figure out the volume.\n\nPV = nRT\n\nwhere P = given pressure - water vapor pressure\nwhere V = your unknown\nwhere n = your total moles of gas\nwhere R = the appropriate Universal Gas Law constant\nwhere T = the temperature in KELVIN
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
236308
The step from pyruvate (CH3COO-) to Acetalaldehyde (CH2CHO) and CO2 is done by the enzyme pyruvate decarboxylase.  The step from Acetalaldehyde to ethanol (CH3CH2OH) is Alcohol dehydrogenase.  I do not have the chemical structures for the enzymes, because they would be way complicated, but I think what you are looking for is.\n\nC6H12O6 + 2ADP + 2 Pi = 2 CH3CH2OH + 2 CO2 + 2ATP + 2H20
802803
The answer is:\n2Al(HCO3)3  -------> Al2(CO3)3  +  3CO2  +  3H20\n\nThe equation is now fully balanced as you have\n2 moles of Al on each side\n6 moles of H on each side\n6 moles of C on each side\n18 moles of O on each side\n\nHope that helps!
890894
C12H22O11+12O2-------12CO2+11H2O+E
301956
The answer is B.\n\nIn your beginning reaction CO2 + C --> CO, you have 2 carbons and 2 oxygens on the reactants side, and you have 1 carbon and 1 oxygen on the products side.\n\nIn order to make the equation balanced, you have to make sure there are the same number of each element on both sides of the equation.\n\nSo to make this chemical equation balanced all you have to do is put 2 as the coefficient of the CO on the products side, making the final equation:  CO2 + C --> 2CO\nThis will give you 2 carbons and 2 oxygens on both sides.
739541
CH4 + 2O2-> CO2 + 2H2O
927756
2H2O2 → 2H20 + 02 is the balanced equation for the above decomposition
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
121351
Are you tryn to have us do your hw!?!?\n\nThe eqn should be in your book.  I dont feel like digging mine out, but its something like;\n\nCp*g*(T2-T1)=energy\n\nSo what u want to find is (t2-t1) Ima call it x\n\n400*50*(x)=800  do the math.  I think thats it.\n\nfor the 2nd question, this one is easy.  The combustion is energy is for 1 mol (per mol).  If you only have .2 moles, the combustion amount is .2* 1370 .
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH