Answer:
p3=0.36atm (partial pressure of NOCl)
Explanation:
2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) Kp = 51
lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively
p1=0.125atm;
p2=0.165atm;
p3=?
Kp=51;
On solving;
p3=0.36atm (partial pressure of NOCl)
Complete Question
Magnesium sulfate forms a hydrate with the formula . What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of
is 120.4 g/mol; H20 = 18 g/mol.
Answer:
The value is of
Explanation:
From the question we are told that
The volume of toluene is
The mass of anhydrous magnesium sulfate is
The formula of the hydrate is
The molar mass of is
From the formula given we see that
1 mole of wil remove 7 moles of
to for the given formula
Hence
120.4 g (1 mole) will remove 7 moles (7 * 18 g = 126 g ) of to for the given formula
Therefore 1 g of x g of
So
=>
From our calculation we obtained that
1 g of will remove
of
Then
of
will remove z g of
of
So
=>
=> of