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2 years ago

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The half life of a radioactive element is 2 million years. a mass of rock originally contained 200g of this element. how many gr

ams of the element are left after 4 million years?

1 answer:

julia-pushkina [17]2 years ago

8 0

Every 2 million years the amount is halved
0 million = 200g
2 million = 200g/2 = 100g
4 million = 100g/2 = 50g

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Which element is chemically similar to chlorine?

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Explanation:

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Which of the following is a physical or chemical characteristic of a mineral?

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2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

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Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

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2 years ago

1. A sample of polystyrene is found to have a number-average molar mass of 89,440 g mol−1 . Neglecting contributions from end gr

Snowcat [4.5K]

Answer:

The weight-average molar mass of polystyrene is 134,160 g/mol.

Explanation:

Molar mass of the monomer styrene , $C_8H_8$ , M=104 g/mol

Given , number average molar mass of the polymer , M'= 89,440 g/mol

Degree of polymerization = n

$n=\frac{M'}{M}=\frac{89,440 g/mol}{104 g/mol}=860$

The weight-average molar mass = $M_{avg}=?$

Molar mass dispersity is ratio of weight-average molar mass to the number average molar mass of the polymer.

$\text{Molar mass dispersity}=\frac{M_{avg}}{M'}$

$1.5=\frac{M_{avg}}{89,440 g/mol}$

$M_{avg}=89,440 g/mol\times 1.5 = 134,160 g/mol$

The weight-average molar mass of polystyrene is 134,160 g/mol.

8 0

2 years ago