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Rasek [7]
3 years ago
8

A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram

s of krypton-81.
If the half-life of krypton-81 is 229,000 years, how much krypton-81 was present when the ice first formed?

Use the formula N = N0 .
Chemistry
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

6.70 grams of krypton-81 was present when the ice first formed

Explanation:

Let use the below formula to find the amount of sample

N= N_0(\frac{1}{2})^n

where

n = \frac{t}{t_{\frac{1}{2}}}

here

t =  458,000 years

t_{\frac{1}{2}} = 229,000

\frac{t}{t_{\frac{1}{2}}} = \\frac{ 458,000}{229,000}

n = \frac{t}{t_{\frac{1}{2}}} = 2.000

Now substituting the values

1.675 = N_0(\frac{1}{2})^{2.000}}

1.675 = N_0\times (0.2500)

N_0= \frac{1.675}{0.2500}

N_0=6.70

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In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.
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Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}

n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{Fe} = 0.5\:mol}

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n_{S} = \dfrac{m_{S}}{MM_{S}}

n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{S} = 0.5\:mol}

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

Fe + S \Longrightarrow FeS

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}

m_{FeS} = n_{FeS}*MM_{FeS}

m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}

\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}

Answer:

44 grams of iron sulfide

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