The reaction between K₂SO₄(aq) and SrI₂(aq) produces KI(aq) and SrSO₄(s) as products.
The reaction is
K₂SO₄(aq) + SrI₂(aq) → KI(aq)+ SrSO₄(s)
To balance the equation both side of the reaction should have same number of atoms in each element.
Right hand side of the reaction has 1 K, 1 I, 1 Sr, 1 S and 4 O atoms while 2 K, 2 I, 1 Sr,1 S and 4 O present in left hand side of the reaction.
Hence, number of I atoms and number of K atoms are not balanced.
To balance the K atoms we should add 2 before KI. Then I atoms will be 2 at the right hand side.
Hence, the balanced reaction equation is
K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s)
Answer:
y1 = 0.3162
y2 = 0.6838
Explanation:
ok let us begin,
first we would be defining the parameters;
at 25°C;
1-propanol P1° = 20.90 Torr
2-propanol P2° = 45.2 Torr
From Raoults law:
P(1-propanol) = P⁰ × X(1-propanol)
P(1-propanol) = 20.9 torr × 0.45 = 9.405
P(1-propanol) = 9.405 torr
Also P(2-propanol) = P⁰ × X(2-propanol)
P(2-propanol) = 45.2 torr × 0.45
P(2-propanol) = 20.34 torr
but the total pressure = sum of individual pressures
total pressure = 9.405 + 20.34
total pressure = 29.745 torr
given that y1 and y2 represent the mole fraction of each in the vapor phase
y1 = P1 / total pressure
y1 = 9.405/29.745
y1 = 0.3162
Since y1 + y2 = 1
y2 = 1 - y1
∴ y2 = 1 - 0.3162
y2 = 0.6838
cheers, i hope this helps.
Answer:
D
Explanation:
P4 is polyatomic molecule
