Answer:
A and B
Explanation:
This is because there was emission of gamma (Y) radiations in both the reactions.
Answer: Oxygen and Carbon
Explanation:
<h3>
Answer:</h3>
5.6 Liters
<h3>
Explanation:</h3>
- N.T.P. refers to the standard temperature and pressure (S.T.P).
We need to know that;
- One mole of a gas occupies a volume of 22.4 liters at N.T.P.
In this case;
We have 11 g of CO₂
But, 1 mole of CO₂ occupies 22.4 l at N.T.P.
1 mole of CO₂ = 44 g
Therefore;
44 g of CO₂ = 22.4 liters
What about 11 g ?
= (11 g × 22.4 l)÷ 44 g
= 5.6 l
Therefore, 11 g of CO₂ will occupy a volume of 5.6 liters at N.T.P.
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: 94.07%
Explanation:
Percentage yield can be calculated by the formula
%yield = Experimental yield/Theoretical yield x100
Experimental yield = 7.93g
Theoretical yield = 8.43
%yield = Experimental yield/Theoretical yield x100
%yield = 7.93/8.43 x 100 = 94.07%
