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3 years ago

7

A sample taken from a layer of mica in a canyon has 2.10 grams of potassium-40. A test reveals it to be 2.6 billion years old. H

ow much potassium-40 was in the sample originally if the half-life of potassium-40 is 1.3 billion years?
A.
4.20 g
B.
8.40 g
C.
12.6 g
D.
16.8 g
E.
25.2 g

1 answer:

Setler79 [48]3 years ago

5 0

Problem One

<em><u>Formula</u></em>

N(t) = No * (1/2)^[t/t_1/2]

<em><u>Givens</u></em>

N(t) = the current mass of the sample = 2.10 grams

No = The original mass of the sample = No [We're trying to find this].

t = time elapsed which is 2.6 billion years or 2.6 * 10^9 years.

t1/2 = the 1/2 life time which is 1.3 billion years of 1.3 *10^9

<em><u>Solution</u></em>

2.10 grams = No (1/2)^(2.6*10^9/1.3 * 10^9)

The 10^9s cancel and you are left with 2.6/1.3 = 2

2.10 grams = No (1/2)^2

2.10 grams = No (1/4) Multiply both sides by 4

2.10 * 4 = No (1/4)*4

8.4 grams = No

which is how many grams you originally had.

Answer B.

Problem Two

$_{1}^{2}\text{H} + _{1}^{2}\text{H} \longrightarrow _{z}^{y}\text{x}+_{0}^{1}\text{n}$

Solve for y

2 + 2 = y + 1

4 = y + 1

y = 3

Solve for z

1 + 1 = z + 0

z = 2

The 2 tells you that it is the second member on the periodic table. That's Helium. So the answer looks like this.

$_{2}^{3}\text{He}$

The mass of the Helium is 3 and its number is two.

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Answer :

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The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

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The expression for solubility constant for this reaction will be,

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$4.2\times 10^{-8}=0.073\times [Br^-]$

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Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgBr\rightleftharpoons Ag^++Br^-$

The expression for solubility constant for this reaction will be,

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In other words, Gay-Lussac's law states that when a gas undergoes a constant volume transformation, the ratio of the pressure exerted by the gas temperature remains constant:

$\frac{P}{T} =k$

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$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

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Hello there!

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Regards!

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