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3 years ago

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A 10 L mixture is made up of 40% antifreeze. How much of the mixture needs to be removed and filled with pure antifreeze to reac

h a concentration of 60% antifreeze

1 answer:

malfutka [58]3 years ago

8 0

Answer:

3.3 liters of the mixture is needed to be removed and filled with 3.33 liters pure antifreeze to reach a concentration of 60% antifreeze.

Explanation:

A 10 L mixture is made up of 40% antifreeze.

Initial Volume of antifreeze in 40% mixture = 40% of 10 L = 4L

Let volume of pure antifreeze added = x

Volume of antifreeze removed = 40% of x= 0.4 x

Volume of antifreeze in 60% mixture = 60% of 10 L = 6 L

Volume of antifreeze left after removal of 0.4 x L of antifreeze and addition of x L of pure antifreeze will be equal to 6 L of antifreeze in the final solution.

4L - (0.4 x ) + x = 6L

x = 3.33 L

3.3 liters of the mixture is needed to be removed and filled with 3.3 liter pure antifreeze to reach a concentration of 60% antifreeze.

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The matter in this series of images is going through a change. Draw a conclusion about the type of change shown. Is it a physica

34kurt

Answer:

This is a chemical change because it has been lit on fire. When it is lit on fire, the fire has been made, heat has been made, and gasses produced from the fire have also been made. Because this introduces new matter to the pictures, it is a chemical reaction.

Explanation:

8 0

3 years ago

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

Brrunno [24]

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:

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

6 0

3 years ago

Which of the following definitions characterizes an element?

Amanda [17]

<span>The answer is A because an elements only consists of one type of atom. Example. the element H (hydrogen) only contains 1 hydrogen atom. I hope this helps!! </span>

6 0

4 years ago

Given that the free energy of the twist-boat conformer of cyclohexane is 5.3 kcal/mol greater than that of the chair conformer,

marishachu [46]

Answer:

0.013%

Yes, it does. The answer agrees with the statement.

Explanation:

Both conformers are in equilibrium, and it can be represented by the equilibrium equation K:

K = [twist-boat]/[chair]

The free energy between them can be calculated by:

ΔG° = -RTlnK

Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C + 273 = 298 K).

ΔG° = 5.3 kcal/mol * 4.182 kJ/kcal = 22.165 kJ/mol = 22165 J/mol

22165 = -8.314*298*lnK

-2477.572lnK = 22165

lnK = -8.946

K = $e^{-8.946}$

K = 1.30x10⁻⁴

[twist-boat]/[chair] = 1.30x10⁻⁴

[twist-boat] = 1.30x10⁻⁴[chair]

The percentage of the twist-boat conformer is:

[twist-boat]/([twist-boat] + [chair]) * 100%

1.30x10⁻⁴[chair]/(1.30x10⁻⁴[chair] + [chair]) *100%

0.013%

The statement about the conformers is that the chair conformer is more stable, and because of that is more present. So, the answer agrees with it.

3 0

3 years ago

How are organelles specialized to perform various tasks in a cell?

aksik [14]

Answer:

Organelles are structures within a cell that perform specific functions like controlling cell growth and producing energy. Plant and animal cells can contain similar types of organelles. However, certain organelles can only be found in plant cells and certain organelles can only be found in animal cells.

8 0

3 years ago