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2 years ago

8

Type the correct answer in the box. a pharmacist works with a 1.75 m solution of sodium bromide (nabr) and water. the volume of

the solution is 84.0 milliliters. if the pharmacist dilutes the solution to 1.00 m, what is the volume of the new solution? express your answer to three significant figures. the volume of the new solution is milliliters.

1 answer:

Elodia [21]2 years ago

8 0

Answer: 147 mL

Explanation:

<u>Given:</u>

Molarity of the sodium bromide (NaBr) solution (M1) = 1.75 M

Volume of the solution (V1) = 84 mL

Molarity of the diluted NaBr solution (M2) = 1 M

Using the dilution formula to solve for V2:

$\begin{gathered}M_{1} V_{1}=M_{2} V_{2} \\V_{2}=(1.75 \mathrm{M} \times 84 \mathrm{~mL}) / 1 \mathrm{M} \\V_{2}=147 \mathrm{~mL}\end{gathered}$

Therefore, the new volume of the solution is 147 mL

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Answer:

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According top the Henry's Law, the concentration of a gas in a liquid is directly proportional to the partial pressure of the gas.

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$P_2$ = final partial pressure of gas = ?

Now put all the given values in the above formula, we get the final partial pressure of the gas.

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4 years ago

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

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To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

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Here, R is radius and is related to a as follows:

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Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

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now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

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$1 pm=10^{-10}cm$

Thus,

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$r_{O^{2-}}=1.4\times 10^{-8}cm$

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Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

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Thus, atomic packing factor is 0.758.

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3 years ago

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