Answer:
0.0036 moles of NaOH are present
Explanation:
A solution of 0.1mol/dm³ contains 0.1 moles of solute (In this case, NaOH), per 1 dm³ of solution.
36cm³ are in dm³ (1000cm³ / 1dm³):
36cm³ * (1dm³/1000cm³) = 0.036dm³
The moles are:
0.036dm³ * (0.1mol/dm³) = 0.0036 moles of NaOH are present
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234
So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left
Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.
1. True
2. False
3. False
4. True
5. True
6. False
7. True
8. True
9. True
10. Base
11. Acid
12. Base
13. Acid
14. Neutral
Cells
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Based on Le Chatelier's principle, if the equilibrium of a given reaction is disturbed through changes in temperature, pressure, volume or concentration, then it will shift in a direction so as to undo the effect of the change induced.
The given reaction is:
2SO2(g) + O2(g) ↔ 2SO3(g)
As per Boyle's law,
V α 1/P
as the volume is increased the pressure of the system would decrease. The equilibrium would then shift in a direction of increased pressure. This would be towards the reactants where there are more moles of the gas.
Therefore, the equilibrium would shift to the left towards the reactants.