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Natasha_Volkova [10]
2 years ago
12

If the half-life of Carbon-14 is 5700 years, how many years would it take a sample to decay from 1 gram to 31.3 mg

Chemistry
1 answer:
andrew-mc [135]2 years ago
3 0

Answer:

28500 years

Explanation:

Applying,

A = A'(2^{x/y})............... Equation 1

Where A = Original mass of Carbon-14, A' = Final mass of carbon-14 after decaying, x = total time, y = half-life.

From the question,

Given: A = 1 g, A' = 31.3 mg = 0.0313 g, y = 5700 years.

Substitute these values into equation 1

1 = 0.0313(2^{x/5700})

2^{x/5700} = 1/0.0313

2^{x/5700}  = 31.95

2^{x/5700} ≈ 32

2^{x/5700} ≈ 2⁵

Equating the base and solve for x

x/5700 ≈ 5

x ≈ 5×5700

x ≈ 28500 years

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16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)

Moles of calcium nitrate = \frac{31.3 g}{164 g/mol}=0.1908 mol

Moles of ammonium fluoride = \frac{38.7 g}{37 g/mol}=1.046 mol

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

\frac{1}{2}\times 1.046 mol=0.523 mol calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

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