Answer:
-205.7kj
Explanation:
Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of the reaction.
The expression for enthalpy for the following reaction will be,
where,
n = number of moles
Now put all the given values in the above expression, we get:
Therefore, the enthalpy of the following reaction is, -205.7kj
I have no idea!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
MM Zn(NO₃)₂ = 189.36 g/mol
mass 1 mol Zn(NO₃)₂ = 189.36 g
mass hydrate = 100 / 63.67 x 189.36 = 297.409 g
mass 1 mol hydrate = 297.409 g
MM hydrate = 297.409 g/mol
MM hydrate = MM Zn(NO₃)₂ + MM xH₂O
297.409 = 189.36 + x(18)
x = 6
What helps me to balance equations is to list the elements i have on each side of the equation, and use tally marks to see what I have and don't have. Then when I'm done balancing, I tally again to make sure everything matches up.
On the left side, you have 1 Al, and 2 O. On the right side, 1 Al and 3 O.
In order for the equation to balance, you need to place a 2 in front of the AlO on the right side. This would make the Al have 2 atoms and the O have six. On the left side, you need to place a 2 in front of the Al and a 3 in front of the O, making it six. Left side: 2 Al's 6 O's. Right side: 2 Al's and 6 O's. Matches!
