Answer:
Faraday's constant will be smaller than it is supposed to be.
Explanation:
If the copper anode was not completely dry when its mass was measured, mass of the copper must be heavier than it should have been. Hence, the calculated Faraday’s constant would be smaller than it is supposed to be since when calculating Faraday’s Constant, the charge transferred is divided by the moles of electrons.
Answer : The activation energy for the reaction is, 119.7 J
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 271 K
= rate constant at 281 K = 
= activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 271 K
= final temperature = 281 K
Now put all the given values in this formula, we get:
![\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{271K}-\frac{1}{281K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B2K_1%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B271K%7D-%5Cfrac%7B1%7D%7B281K%7D%5D)
Therefore, the activation energy for the reaction is, 119.7 J
Answer:
How much energy would be produced from the reaction of 2.40 moles hydrogen with 3.95 moles chlorine?
Explanation:
Given,
2.40 moles hydrogen reacts with 3.95 moles chlorine.
From the balanced chemical equation,
1mol. of H2 reacts with 1mol. of Cl2
then,
2.40mol. of H2 reacts with ----?mol of Cl2
So, the remaining moles of Cl2 is in excess.
The limiting reagent is--- H2.
1mol. of H2 releases --- 554kJ of energy
then,
2.40mol of H2 releases ---- ?
Answer is: deltaH =-1329.6kJ
I think the chemical reaction is:<span>
N2H4 + 2 H2O2-> N2 + 4H2O
We are given the amount reactants allowed to react. This will be the starting point of the reaction. First, is to find the limiting reactant.
9.24 g H2O2 ( 1 mol / 34.02 g ) = 0.27 mol H2O2
6.56 g N2H4 ( 1mol / 32.06) = 0.20 mol N2H4
Since from the reaction we have 1:2 ratio of the reactants then the limiting reactant is hydrogen peroxide. We will use this to find the amount of N2 produced.
0.27 mol H2O2 ( 1 mol N2 / 2 mol H2O2 ) ( 14.01 g N2 / 1 mol N2) =1.89 g N2
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