Answer:
See explanation.
Explanation:
Hello!
In this case, since we know the heat of reaction per gram of reactant and we should know the total energy of reaction, but it is not there, we are going to assume it is 1200 J as usual in these problems, so you can change it to whatever your given heat is.
In such a way, we set up the math as shown below:

Which results:

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Answer:
No. Of Moles of zinc = m/Ar
= 13/ 65.38 = 0.198 moles
From balanced equation, Mole ration between CuSO4 and Zn is 1 : 1
So only 0.198 moles of CuSO4 reacts, it is in excess
Mass = no of Moles X Mr
Mass = 0.198 X 159.5 = 31.59 grams
Volume = mass m denisty
Volume j 31.59 / 3.6 = 8.78 ml
Explanation:
i think this wrong
Answer:
At equilibrium, the concentration of
is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+
⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no
nor
present. Immediately,
and
are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: [
]=0 ; [
]= 0 ; [
]=0.60M
C: [
]=+x ; [
]= +x ; [
]=-2x
E: [
]=0+x ; [
]= 0+x ; [
]=0.60-2x
Now we can use the constant information:
![K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5Bproducts%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D%7B%5Breactants%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D)
= 
= 
= 




At equilibrium, the concentration of
is going to be 0.30M
I’m pretty sure it’s B
Hope that helps