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yaroslaw [1]
3 years ago
8

Phosphorus-32 has a half life of 14.3 days. How many days will it take for a

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
3 0

It will take 42.9 days

<h3>Further explanation</h3>

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

t1/2=14.3 days

Nt/No=1/8

\tt Nt=No(\dfrac{1}{2})^{T/t1/2}\\\\\dfrac{Nt}{No}=\dfrac{1}{2}^{T/t1/2}\\\\\dfrac{1}{8}=\dfrac{1}{2}^{T/t1/2}\\\\(\dfrac{1}{2})^3=\dfrac{1}{2}^{T/t1/2}\\\\3=T/14.3\rightarrow T=42.9~days

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Answer:

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

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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 :

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Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

\frac{2}{1}\times 0.1908 mol=0.3816 molof dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

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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.

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Answer: Three things are missing:


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