Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Answer:
V2 = 3.11 x 105 liters
Explanation:
Initial Volume, V1 = 2.16 x 105 liters
Initial Temperature, T1 = 295 K
Final Temperature, T2 = 425 K
Final Volume, V2 = ?
These quantities are related by charle's law and the equation of the law is given as;
V1 / T1 = V2 / T2
V2 = T2 * V1 / T1
V2 = 425 * 2.16 x 105 / 295
V2 = 3.11 x 105 liters
Hydrogen + oxygen --> water
2,1g + 16,8g = x
x = 18,9g
