Answer:
3 years
Explanation:
Given data:
Initial amount of sample = 160 Kg
Amount left after 12 years = 10 Kg
Half life = ?
Solution:
at time zero = 160 Kg
1st half life = 160/2 = 80 kg
2nd half life = 80/2 = 40 kg
3rd half life = 40 / 2 = 20 kg
4th half life = 20 / 2 = 10 kg
Half life:
HL = elapsed time / half life
12 years / 4 = 3 years
Answer:
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Answer:
12.50g
Explanation:
T½ = 2.5years
No = 100g
N = ?
Time (T) = 7.5 years
To solve this question, we'll have to find the disintegration constant λ first
T½ = In2 / λ
T½ = 0.693 / λ
λ = 0.693 / 2.5
λ = 0.2772
In(N/No) = -λt
N = No* e^-λt
N = 100 * e^-(0.2772*7.5)
N = 100*e^-2.079
N = 100 * 0.125
N = 12.50g
The sample remaining after 7.5 years is 12.50g
There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
E = mct
Energy = (mass) x (specific heat capacity of water) x (change in temp)
585.24 = 53.2 x 4.2 x (X-24.15)
585.24 divided by 53.2 divided by 4.2 = X - 24.15
2.62 = X - 24.15
X= 26.77degrees C
(Specific heat capacity for water is 4.2 but is different for other liquids)