I really need help with this!
1 answer:
You might be interested in
Answer:
Explanation:
count given by old sample = .97 disintegrations per minute per gram
count given by fresh sample = 6.68 disintegrations per minute per gram
Half life of radioactive carbon = 5568 years
rate of disintegration
dN / dt = λ N
In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .
Let initial no of radioactive be N₀ and after time t , number reduces to N
N₀ / N = 6.68 / .97
Now
λ is disintegration constant
λ = .693 / half life
= .693 / 5568
= .00012446 year⁻¹
Putting the values in the equation above
1.929577 = .00012446 t
t = 15503.6 years .
<u>Answer:</u>
<u>For A:</u> The mass of oxygen per gram of sulfur for sulfur dioxide is 0.997 g
<u>For B:</u> The mass of oxygen per gram of sulfur for sulfur trioxide is 1.5 g
<u>Explanation:</u>
- <u>For A:</u>
The chemical equation for decomposition of sulfur dioxide follows:
We are given:
Mass of sulfur = 3.47 g
Mass of oxygen = 3.46 g
To calculate the mass of oxygen per gram of sulfur, we apply unitary method:
For every 3.47 g of sulfur, 3.46 g of oxygen will form.
So, for 1 gram of sulfur, of oxygen will form.
Hence, the mass of oxygen per gram of sulfur for sulfur dioxide is 0.997 g
- <u>For B:</u>
The chemical equation for decomposition of sulfur trioxide follows:
We are given:
Mass of sulfur = 5.50 g
Mass of oxygen = 8.25 g
To calculate the mass of oxygen per gram of sulfur, we apply unitary method:
For every 5.50 g of sulfur, 8.25 g of oxygen will form.
So, for 1 gram of sulfur, of oxygen will form.
Hence, the mass of oxygen per gram of sulfur for sulfur trioxide is 1.5 g