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April 24, 2014

April 24, 2014

Posted by **hannah** on Tuesday, November 6, 2012 at 10:30am.

N2H4(l) + O2(g) N2(g) + 2 H2O(l) ÄH = -622.2 kJ

H2(g) + 1/2 O2(g) H2O(l) ÄH = -258.5 kJ

H2(g) + O2(g) H2O2(l) ÄH = -187.8 kJ

Use this information to calculate the enthalpy change for the reaction shown below.

N2H4(l) + 2 H2O2(l) N2(g) + 4 H2O(l) ÄH = ?

i really do0nt know how to do these

- chem- i reallyneed help -
**DrBob222**, Tuesday, November 6, 2012 at 5:17pmAdd equn 1 as is, to 2x equation 2, to 2x the reverse of equation 3 to obtain the equation you want. Then add the delta H values. When you multiply an equation you must multiply delta H values too. When reversing an equation, change the sign of delta H.

- chem- i reallyneed help -
**rebekah**, Tuesday, November 6, 2012 at 6:30pmso is it - 764.2?

- chem- i reallyneed help -
**hannah**, Tuesday, November 6, 2012 at 8:35pmoops thats me ^

- chem- i reallyneed help -
**DrBob222**, Wednesday, November 7, 2012 at 12:41pm763.6 is what I get.

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