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 is: the percent composition of Hg in the compound is 71.5%.
Balanced chemical reaction: Hg + Br₂ → HgBr₂.
m(Hg) = 60.2 g; mass of the mercury.
m(Br₂) = 24.0; mass of the bromine.
m(HgBr₂) = m(Hg) + m(Br₂).
m(HgBr₂) = 60.2 g + 24 g.
m(HgBr₂) = 84.2 g; mass of the compound.
ω(Hg) = m(Hg) ÷ m(HgBr₂) · 100%.
ω(Hg) = 60.2 g ÷ 84.2 g · 100%.
ω(Hg) = 71.5%.
NH3 has 1 Nitrogen atom and 3 hydrogen atoms.
LIKE DISSOLVES LIKE. Since Ccl4 is non-polar, it'll be soluble in any non-polar solvent. Hope this helps you!
The columns of the periodic table, also referred to as "groups" contain elements with similar reactive properties, due to these elements having a similar configuration of electrons in their outer shell.