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
Hey there!
The correct answer to your question is: Spectrum
A spectrum, by definition, is a range of colors which are produced by separating certain colors of light.
Thank you!
Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
Answer:
The main purpose for converting numbers into scientific notation is to make calculations with surpisingly large or small numbers less complicated. Since zeros are not used to place the decimal point, all of the digits in a number in scientific notation are meaningful and easier to read. Hope this helps!
Explanation:
