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3 years ago

Q2.

Chemistry

1 answer:

Serga [27]3 years ago

4 0

Answer:

266 liters Carbon iv oxide will be formed

Explanation:

The first thing we need to do here is to calculate the number of moles of oxygen reacted.

This can be calculated using the ideal gas equation;

Mathematically;

PV = nRT

From the question;

P = pressure = 10.5 atm

V = volume = 400 liters

n = number of moles = ?

T = temperature = 125 + 273 = 398K

R = molar gas constant = 0.0821 L•atm•K^-1•mol^-1

Substituting these values, we have ;

n = PV/RT

n = (10.25 * 400)/(398 * 0.0821) = 125 moles

From the question;

15 moles oxygen gave 10 moles CO2

125 moles oxygen will give x moles CO2

x = (125 * 10)/15 = 83 moles

Now, we want to know the volume of CO2, present in 83 moles using the given reaction conditions.

Mathematically;

PV = nRT

V = nRT/P

= (83 * 0.0821 * 398)/10.25 = 265.65 which is approximately 266 Liters

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Light reaction do not depends on dark reaction because the reaction that occur in presence of light cannot occur in dark . Photolysis, one of most important light reaction cannot take place in dark.

Explanation:

If concentration of carbon dioxide is decreased by 50% then the oxygen production will not be affected because oxygen is produced from the reduction of water.

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Lake Eola is a sinkhole lake. What reasoning best describes how it formed?

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NemiM [27]

Answer:

Explanation:

H₂SO₄:

Element (symbol) No of atoms in No of compounds Total atoms

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Element (symbol) No of atoms in No of compounds Total atoms

1 molecule coefficient

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3 years ago

NEED HELP

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Answer:

Explanation:

V1/T1 =V2/T2 at constnant pressure

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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

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

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