The answer & explanation for this question is given in the attachment below.
A kinetics experiment is set up to collect the gas that is generated when a sample of chalk, consisting primarily of solid CaCO3
1 answer:
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Answer : The correct option is, (a) 0.44
Explanation :
First we have to calculate the concentration of .
Now we have to calculate the dissociated concentration of .
The balanced equilibrium reaction is,
Initial conc. 1.0 M 0
At eqm. conc. (1.0-x) M (2x) M
As we are given,
The percent of dissociation of =
= 28.0 %
So, the dissociate concentration of =
The value of x = = 0.28 M
Now we have to calculate the concentration of at equilibrium.
Concentration of = 1.0 - x = 1.0 - 0.28 = 0.72 M
Concentration of = 2x = 2 × 0.28 = 0.56 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be:
Now put all the values in this expression, we get :
Therefore, the equilibrium constant for the reaction is, 0.44
Answer: 32.94 g
Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:
From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.
for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm
V = 15.0 L
T = 350.8 + 273 = 623.8 K
For krypton,
n = 0.146 moles
for chlorine,
n = 0.439
From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.
Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.
So, the amount of product formed is calculated from moles of krypton.
Molar mass of krypton tetrachloride is 225.61 gram per mol.
There is 1:1 mol ratio between krypton and krypton tetrachloride.
= 32.94 g of
So, 32.94 g of the product will form.
Answer:
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Explanation:
Volume of NaOH = 1.7 ml = 0.0017 L
Molarity of NaOH = 0.0811 M
Moles of NaOH = n
n = 0.0001378 mol
According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.
Then 0.0001378 mol of NaOH will neutralize:
of sulfuric acid.
Concentration of sulfuric acid in the acid rain sample: x
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.