From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ
<h3>What is Hess' law?</h3>
Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.
From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:
A---> C = A ---> B + B ---> C
ΔH of A---> C = 30 kJ + 60 kJ
ΔH = 90 kJ
Therefore, the enthalpy of the reaction A---> C is +90 kJ
The above reaction A---> C can be shown in the enthalpy diagram below:
A -------------------> C (ΔH = +90 kJ)
\ /
\ / (ΔH = +60 kJ)
(ΔH = +30 J) \ /
> B
Learn more about enthalpy and Hess law at: brainly.com/question/9328637
These problems are a bit interesting. :)
First let's write the molecular formula for ammonium carbonate.
NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)
17.6 gNH4CO3
Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.
17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)
Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)
NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol
17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4
Now just take the molar mass we found to convert that amount into moles!
4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4
1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)
Ksp{Zn(OH)₂}=1.2*10⁻¹⁷
2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)
Ksp{Zn(CN)₂}=2.6*10⁻¹³
Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}
Zn(OH)₂ precipitates first
H=(2x1.008)=2.016
O= 15.999
15.999
+ 2.016
______
18.015 g/mol :)
