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3 years ago

What process is used to refine aluminum from its ore bauxite?

Chemistry

1 answer:

Murljashka [212]3 years ago

7 0

Answer:

D. Smelting

Explanation:

Bauxite is the aluminum ore from which alumina, also known as aluminum oxide, Al₂O₃, is produced. Bauxite is extracted from the topsoil regions of some subtropical and tropical regions, and the Bayer process is primarily then used to produce alumina from the bauxite.

Aluminum is produced from the alumina by an aluminum smelting process known as the Hall—Heroult electrolytic process which involves the use of a carbon anode and direct current to produce aluminum by reducing the aluminum oxide

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What is the total pressure of a gas mixture containing partial pressures of

Advocard [28]

Answer:

1.54 atm

Explanation:

By Dalton's Law Of partial pressure,

Total Pressure = Sum of all partial pressures

So,P= P1 + P2 + P3

Therefore, P=0.23+0.42+0.89

=1.54 atm

8 0

3 years ago

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w

Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619= [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0

3 years ago

A standard 1.00 kg mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 2.50

Elden [556K]

Answer:

The section of the bar is 2.92 inches.

Explanation:

Mass of the steel cut ,m = 1.00 kg = 1000 g

Volume of the steel bar = V = Area × height

Height of the of the section of bar = h

Area of Equilateral triangular = $\frac{\sqrt{3}}{4}a^2$

a = 2.50 inches

Cross sectional area of the steel mass = A

$A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2$

$V = 2.71 inches^2\times h$

Density of the steel = d = $7.70 g/cm^3$

$1cm^3 = 0.0610237 inches^3$

$d=\frac{m}{v}$

$\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}$

$h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}$

h = 2.92 inches

The section of the bar is 2.92 inches.

3 0

3 years ago

A sample of a compound contains 60.0 g C and 5.05 g H. Its molar mass is 78.12 g/mol. What is the compound’s molecular formula?

Inessa [10]

A sample of a compound contains 60.0 g C and 5.05 g H.

divide by molar mass of C(12) and H(1) to get molar ratio

C: 60/12=5 and H: 5/1=5

so C:H=5:5=1:1

total molar mass=78

divide by 1C+1H to find the formula: 78/(12+1)=78/13=6

compound is C6H6

5 0

4 years ago

Read 2 more answers

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago