Answer:
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Explanation:
Step 1: Data given
Kp = 4.7 x 10^3 at 400K
Pressure of CH3OH = 0.250 atm
Pressure of HCl = 0.600 atm
Volume = 10.00 L
Step 2: The balanced equation
CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)
Step 3: The initial pressure
p(CH3OH) = 0.250atm
p(HCl) = 0.600 atm
p(CH3Cl)= 0 atm
p(H2O) = 0 atm
Step 3: Calculate the pressure at the equilibrium
p(CH3OH) = 0.250 - X atm
p(HCl) = 0.600 - X atm
p(CH3Cl)= X atm
p(H2O) = X atm
Step 4: Calculate Kp
Kp = (pHO * pCH3Cl) / (pCH3* pHCl)
4.7 * 10³ = X² /(0.250-X)(0.600-X)
X = 0.249962
p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm
p(HCl) = 0.600 - 0.249962 = 0.350038 atm
p(CH3Cl)= 0.249962 atm
p(H2O) = 0.249962 atm
Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)
Kp = 4.7 *10³
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Answer:
Explanation:
Approx.
425
⋅
g
Explanation:
2
A
l
(
s
)
+
3
C
l
2
(
g
)
→
2
A
l
C
l
3
(
s
)
You have given a stoichiometrically balanced equation, so bravo.
The equation explicitly tells us that
54
⋅
g
of aluminum metal reacts with
6
×
35.45
⋅
g
C
l
2
gas to give
266.7
⋅
g
of
aluminum trichloride
hope this helps
Answer:
less than 5 centimeters because it is a nonvascular plant
Explanation:
Mosses are a group of plants under the division Bryophyta. They are said to be the most primitive plant life in existence as they lack true roots, stems and leaves. They also lack vascular system, hence, they are regarded as non-vascular plants. They usually grow in very small sizes (about 0.2 - 10cm).
According to this question, Penny bought a club moss plant for her water garden and needs to know how tall the plant will grow so she know how much space it will need. Since it is a miss plant that lacks vascular tissues i.e. nonvascular, it will likely grow less than 5 centimeters in height.
Wouldn't you take Avogadro' number and multiply by 3.80 maybe
Answer:
118750 ml
Explanation:
The chemical equation for complete combustion of hexane is given as;
2C6H14 + 19O2 → 12CO2 + 14H2O
From the equation of the reaction;
2 mol of C6H14 reacts with 19 mol of O2
2 ml of C6H14 reacts with 19 ml of O2
2500 mL of C6H14 would react with x ml of O2
2 = 19
2500 = x
x = 2500 * 19 / 2 = 23750 ml
Since oxygen is 20% of air;
23750 = 20 / 100 * (Volume of air)
Volume of air = 23750 * 100 / 20 = 118750 ml
