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2 years ago

Calculate the density of CO2 at a pressure of 685.0 torr and 41.0°C . R=0.0821 (Latm)/(mol K)

Chemistry

1 answer:

Keith_Richards [23]2 years ago

5 0

T=41+273=314 k

M=(12)+(16×2)=44g/mol

d=PM/RT

d=685×44/0.0821×314

d=1169.15 g/L

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What is the density of a liquid that has a mass of 27g and a volume of 30mL?

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Answer: 7 g/cm to the third power
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3 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

IRINA_888 [86]

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

Area is calculated as:

$A=6R^{2}\sqrt{3}$

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

$R=\frac{a}{2}$

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

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

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.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

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

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

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

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

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,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

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3 years ago

Read 2 more answers

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Its known as covalently bonded atoms

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2 years ago

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garri49 [273]

Answer:

Mg would blow off. AI would be affective to copper but not to MG

Explanation:

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2 years ago

What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?

Varvara68 [4.7K]

The dissociation of formic acid is:

$HCOOH \rightleftharpoons HCOO^{-} + H^{+}$

The acid dissociation constant of formic acid, $k_a$ is:

$k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}$

Rearranging the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

pH = 2.75

$pH = -log[H^{+}]$

$[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}$

$pk_a = 3.75$

$k_a = 10^{-3.75} = 1.78\times 10^{-4}$

Substituting the values in the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}$

Hence, the ratio is $\frac{1}{10}$ .

4 0

3 years ago

Calculate the density of CO2 at a pressure of 685.0 torr and 41.0°C . R=0.0821 (L*atm)/(mol *K)

Calculate the density of CO2 at a pressure of 685.0 torr and 41.0°C . R=0.0821 (Latm)/(mol K)