The complete question is as follows: The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) CH4(g) + CCl4(g) Calculate the equilibrium partial pressures of all species when CH2Cl2(g) is introduced into an evacuated flask at a pressure of 0.968 atm at 350 K. PCH2Cl2 = atm PCH4 = atm PCCl4 = atm
Answer: The equilibrium partial pressures of all species, that is,
,
and
is 0.420 atm, 0.420 atm and 0.128 atm.
Explanation:
For the given reaction equation, the initial and equilibrium concentration of involved species is as follows.
Initial: 0.968 atm 0 0
Equilibrium: (0.968 - 2x) x x
Now,
for this reaction is as follows.


Thus, we can conclude that the equilibrium partial pressures of all species, that is,
,
and
is 0.420 atm, 0.420 atm and 0.128 atm.
First, you write the balanced equation for the reaction
2HCl + Mg(OH)2 ----->MgCl2 + 2H2O
Next, find the number of moles of Magnesium Hydroxide
Number of moles of Mg(OH)2 =
mass / Mr = 3.33 / 24 +2(1+16)
=0.0574 mol
Find the mole ratio between HCl and Mg(OH)2 which is 1:2
Number of moles of HCl = 0.0574 x 2 = 0.1148 mol
Mass of HCl = Mr x no. of moles
=0.1148 x (1+35.5)
=4.19 g
<span>moles glucose = 19 g / 180 g/mol= 0.105
M = 0.105 / 0.100 L = 1.05
moles in 20.0 mL = 1.05 M x 0.0200 L = 0.0216
New concentration = 0.0216 /0.500 L = 0.0432 M
moles in 100 mL = 0.100 L x 0.0432 = 0.00432
mass = 0.00432 x 180 g/mol= 0.778 g</span>
I believe the answer would be C
Answer:
a) 37.04% b) 37.04% c) 9.63%
Explanation:
The theoretical percent recovery (Tr), is the total percentage of each compound in the sample. Depending on the technique used to recovery the compounds, the percent recovery will be less than the theoretical, because no technique is 100% efficient.
So, to calculate the theoretical, it will be the mass of the compound divided by the mass of the sample multiplied by 100%.
a) Tr = (250 mg)/(675 mg) * 100%
Tr = 37.04%
b) Tr = (250 mg)/(675 mg) * 100%
Tr = 37.04%
c) Tr = (65 mg)/(675 mg) * 100%
Tr = 9.63%