Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
I think the correct answers from the choices listed above are the first, third and the last option. Ionic compounds are compounds that dissociates into ions when in aqueous solution. From the list, NH4Cl, KF and MgO are the ionic compounds. Hope this answers the question.
Answer:
1.40*10⁻² M
Explanation:
We have the solubility formula
Solubility,
S = KH*P
where
KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm
P = atmospheric pressure = 0.400 atm
Hence, we have
S = KH*P
= (3.50*10⁻² mol/L.atm)*(0.400 atm)
= 1.40*10⁻² mol/L
But 1 mol/L = 1 M,
Hence, the answer (1.40*10⁻² mol/L
) is equivalent to
= 1.40*10⁻² M
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The answers will be 21.5 L N2
