Answer:
See Explanation
Explanation:
For SF6;
Since;
1.25 g of S corresponds to 4.44g of F
1 g of sulphur corresponds to 1 * 4.44/1.25 = 3.55
For SF4;
Since;
1.88 g of S corresponds to 4.44g of F
1 g of sulphur corresponds to 1 * 4.44/ 1.88 = 2.36
Hence;
Mass of oxygen per gram of sulphur in SF6/Mass of oxygen per gram of sulphur in SF4
=
3.55/2.36 = 1.5
Hence the law of multiple proportion is obeyed here.
The rate equation is given as:
k = A e^(- Ea / RT)
Dividing state 1 and state 2:
k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)
k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]
k1/k2 = e^[- Ea / RT1 + Ea / RT2)]
Taking the ln of both sides:
ln (k1/k2) = - Ea / RT1 + Ea / RT2
ln (k1/k2) = - Ea / R (1/T1 - 1/T2)
Since k2 = 4k1, therefore k1/k2 = ¼
ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273
K – 1/ T2)
2.058 x 10^-4 = 1/273 – 1/T2
T2 = 289.25 K
Answer:
1.28 g
Explanation:
Mass of anhydrous compound/molar mass of anhydrous compound = mass of hydrated compound/ molar mass of hydrated compound
Mass of anhydrous compound = ?
Mass of hydrated compound = 2g
Molar mass of anhydrous compound= 160 g/mol
Molar mass of hydrated compound = 250 g/mol
x/160 = 2/250
250x = 2 ×160
x= 2 × 160/250
x= 1.28 g
How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
