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grade 12 chem question: enthalpy change?
the enthalpy of combustion of hydrogen is 243 kj :
H2(g) + 1/2 O2(g) = H20 (g) + 243 KJ
to produce 973 kj, how many grams of hydrogen is needed?
what i did was i got 243 KJ/mol. then divided 973/243 to get 4 mols. then multiplied that by 2.02 g/mol to get 8.08 grams.....is that right?
2 Answers
- Anonymous9 years agoFavorite Answer
Sibyl ... you sure are right ! Congratulations !
- ?Lv 45 years ago
the foremost to this variety of challenge is to multiply the extremely some equations you're given with the help of a perfect quantity, write them backwards or forwards as you go with for, then upload all of them as much as get the needed equation with not something left over. ultimately upload up each and all the kJ's. whenever you multiply an equation with the help of two, 3,..., you multiply the kJ's too. whenever you write an equation backwards, -kJ's substitute into +kJ's. with the help of ways, the warmth of formation of P4O10 is P4 + 5O2 ===> P4O10 -2967kJ. the warmth of formation of PCl5 is P + 5/2Cl2 ===> PCl5 --84kJ. First write: P4O10 ===> P4 + 5O2 +2967 That gets you P4O10 on the left. next upload to that: 6PCl5 ===> 6P + 15Cl2 +504 (greater beneficial with the help of 6 and written bacwards) That gets you 6PCl5 on the left. next upload: 10PCl3 + 5O2 ===> 10Cl3PO -2860kJ (x 10) Now you have 10Cl3PO on the splendid. next comes: 10P4 + 15Cl2 ===> 10PCl3 -3065kJ (x 2.5) That cancels out the P and Cl2 on the splendid and the PCl3 on the left. +2967 + 504 - 2860 - 3065 = -2454kJ, this is the respond.