Answer: -7.4 kJ
Explanation:
The balanced chemical reaction is:
![Fe(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2(g)](https://tex.z-dn.net/?f=Fe%28s%29%2B2HCl%28aq%29%5Crightarrow%20FeCl_2%28s%29%2BH_2%28g%29)
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{FeCl_2}\times \Delta H_{FeCl_2})+(n_{H_2}\times \Delta H_{H_2})]-[(n_{Fe}\times \Delta H_{Fe})+(n_{HCl}\times \Delta H_{HCl})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BFeCl_2%7D%5Ctimes%20%5CDelta%20H_%7BFeCl_2%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20%5CDelta%20H_%7BH_2%7D%29%5D-%5B%28n_%7BFe%7D%5Ctimes%20%5CDelta%20H_%7BFe%7D%29%2B%28n_%7BHCl%7D%5Ctimes%20%5CDelta%20H_%7BHCl%7D%29%5D)
where,
n = number of moles
and
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![\Delta H=[(1\times -341.8)+(1\times 0)]-[(1\times 0)+(2\times -167.2)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-341.8%29%2B%281%5Ctimes%200%29%5D-%5B%281%5Ctimes%200%29%2B%282%5Ctimes%20-167.2%29%5D)
![\Delta H=-7.4kJ](https://tex.z-dn.net/?f=%5CDelta%20H%3D-7.4kJ)
Therefore, the enthalpy change for this reaction is -7.4 kJ
Answer:
Volume of dry gas at STP = 0.432 liters or 432 ml
Explanation:
Given:
Pressure (P) = 740 mmHg - 24 mmHg = 716 mmHg
Temperature (t) = 25 degrees C + 273 K = 298 K
500 ml = 0.5 l
Find:
Volume of dry gas at STP
Computation:
[P1][V1] / T1 = [P2][V2] / T2
[716][0.5] / 298 K = [760][ x Liters] / 273 K
x = 0.432 Liters
Volume of dry gas at STP = 0.432 liters or 432 ml
Answer:
If the half-life of radon-222 is 3.82 days, then how much of a 10.0 gram sample of radon-222 would be left after 7.64 days
✓ 2.50g
Explanation:
Based on the reactants and products,
- The molecular equation of the reaction is: Al(NO3)3 + 3 KOH ---> Al(OH)3 + NaNO3
- The complete ionic equation of the reaction is: Al3+ 3NO3- + 3 K+ 3OH- ---> Al(OH)3(s) + 3 Na+ 3 NO3+
- The net ionic equation of the reaction is:Al3+ 3OH- ---> Al(OH)3(s)
- The spectator ions are: Na+ and NO3+
<h3>What is a precipitate?</h3>
A precipitate is a solid product formed in when two solutions of salts react by double replacement.
The molecular equation of the reaction of Al(NO3)3 solution with a KOH solution is:
Al(NO3)3 + 3 KOH ---> Al(OH)3 + NaNO3
The complete ionic equation of the reaction of Al(NO3)3 solution with a KOH solution is:
Al3+ 3NO3- + 3 K+ 3OH- ---> Al(OH)3(s) + 3 Na+ 3 NO3+
The net ionic equation of the reaction of Al(NO3)3 solution with a KOH solution is:
- Al3+ 3OH- ---> Al(OH)3(s)
- The spectator ions are: Na+ and NO3+
Learn more about ionic equation at: brainly.com/question/25604204