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3 years ago

7. If the pOH of an RbOH solution is 6.32, what is the concentration (molarity) of the base?

Chemistry

1 answer:

liubo4ka [24]3 years ago

6 0

4.79 x 10⁻⁷moldm⁻³

Explanation:

Given parameters:

pOH of RbOH = 6.32

Unknown:

Molarity of the base = ?

Solution:

The pH or pOH scale is used for expressing the level of acidity alkalinity of aqueous solutions.

pOH = -log[OH⁻]

we know the pOH to be 6.32

6.32 = -log[OH⁻]

[OH⁻] = inverse log₁₀(6.32)

[OH⁻] = 4.79 x 10⁻⁷moldm⁻³

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3 years ago

if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius

Lady_Fox [76]

The mass of ore required is 21 700 t.

r = 750 cm

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3} \pi (750 cm)^{3}$ = 1.767 × 10⁹ cm³

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6 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

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3 years ago

A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti

Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)$

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

$n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2$

In such a way, the yielded moles of hydrobromic acid and chlorine are:

$n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2$

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

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3 years ago

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Alexandra [31]

Answer:

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3 years ago

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