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:
The formula for tin(IV) sulfide is SnS
I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ?
Well the formula is :

(where Q is heat, m is mass, c is specific heat and

is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature
Answer:
The first row of elements fits in period <u>6</u>, after the element <u>lanthanum (La)</u>. The second row of elements fits in period <u>7</u>, after the element <u>actinium (Ac). </u>
I hope this helps!
Answer:
The volume you need to transfer from the stock solution is 0.145 l
Explanation:
Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:
number of moles in volume to transfer = number of moles in the final solution
Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:
Ci * Vi = Cf * Vf
where:
Ci = concentration of the stock solution.
Vi = volume of the stock solution to be transferred.
Cf = concentration of the final solution
Vf = volume of the final solution
Then, replacing with the data:
518 mM * Vi = 16.7 mM * 4.5 l
Vi = 16.7 mM * 4.5 l / 518 mM
<u>Vi = 0.145 l or 145 ml</u>
Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.