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3 years ago

What change will be caused by addition of a small amount of hcl to a solution containing fluoride ions and hydrogen fluoride?

Chemistry

1 answer:

agasfer [191]3 years ago

8 0

Answer:

The addition of a small amount of HCl to a solution containing fluoride ions and hydrogen fluoride will cause the equilibrium shift to the production of more hydrogen fluoride.

Explanation:

The solution containing fluoride ions and hydrogen fluoride (a weak acid) may be chemically represented by this equilibrium equation:

F⁻ (aq) + H⁺ (aq) ⇄ HF (aq)

The HCl, a strong acid, added will ionize in water according to this chemical equation:

HCl (aq) → H⁺ (aq) + Cl⁻ (aq)

Then, following Le Chatelir's principle, the addtion of H⁺ ions coming from the HCl dissociation, will increase the concentration of H⁺ in the solution, driving to the consumption of some F⁻ ions, and the production of more HF acid. This is a shift of the equilbrium toward the HF side.

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Calculate the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium

Sergeu [11.5K]

Answer:

1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.

Explanation:

Given data:

mass of cadmium = 6.35 g

Number of atoms of aluminum as 6.35 g cadmium contain = ?

Solution:

Number of moles of cadmium = 6.35 g/ 112.4 g/mol

Number of moles of cadmium = 0.06 mol

Number of atoms of cadmium:

1 mole = 6.022×10²³ atoms of cadmium

0.06 mol × 6.022×10²³ atoms of cadmium/ 1mol

0.36×10²³ atoms of cadmium

Number of atoms of Al:

Number of atoms of Al = 0.36×10²³ atoms

1 mole = 6.022×10²³ atoms

0.36×10²³ atoms × 1 mol /6.022×10²³ atoms

0.06 moles

Mass of aluminum:

Number of moles = mass/molar mass

0.06 mol = m/ 27 g/mol

m = 0.06 mol ×27 g/mol

m = 1.62 g

Thus, 1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.

6 0

3 years ago

Be sure to answer all parts.Consider the following equilibrium process at 686°C:CO2(g) + H2(g) ⇌ CO(g) + H2O(g)The equilibrium c

asambeis [7]

Answer:

a) Kc = 0.578

b) [CO2] = 0.4764 M

[H2]= 0.0164 M

[CO]= 0.0756 M

[H2O]=0.0596 M

Explanation:

Step 1: Data given

Temperature = 686 °C

The equilibrium concentrations of the reacting species are:

[CO] = 0.0520 M

[H2] = 0.0400 M

[CO2] = 0.0810 M

[H2O] = 0.0360 M

Step 2: The balanced equation

CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

For 1 mol CO2 we need 1 mol H2 to produce 1 mol CO and 1 mol H2O

Step 3: Calculate Kc

Kc = [CO][H2O] / [CO2][H2]

Kc = (0.0520 * 0.0360) / (0.0810 * 0.0400)

Kc = 0.578

b) If we add CO2 to increase its concentration to 0.500 mol / L, what will the concentrations of all the gases be when equilibrium is reestablished?

adding CO2 the equilibrium will shift on the right:

Concentrations at the equilibrium will be:

[CO2] = 0.500 -x

[H2]= 0.0400 -x

[CO]= 0.0520+x

[H2O]= 0.0360+x

0.578 = ( 0.0520+x)( 0.0360+x)/ ( 0.500-x)( 0.0400-x)

0.578 = ( 0.0520+x)( 0.0360+x)/ (0.02 -0.500x -0.0400x +x²)

0.01156 - 0.26588 + 0.578x² = 0.001872 + 0.0880x + x²

x = 0.0236

[CO2] = 0.500 - 0.0236 = 0.4764 M

[H2]= 0.0400 - 0.0236 = 0.0164 M

[CO]= 0.0520+0.0236 = 0.0756 M

[H2O]= 0.0360+ 0.0236 = 0.0596 M

5 0

4 years ago

Separate this redox reaction into its balanced component half‑reactions. Use the symbol e− for an electron. Cl2+2Li⟶2LiCl

ahrayia [7]

Answer:

Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation

2Li + 2e⁻ ⟶ 2Li⁺ Reduction

Explanation:

The question requests to split the equation below into half equations;

Cl2+2Li⟶2LiCl

In redox chemistry, splitting into half equations simply means highlighting the reduction and oxidation reactions of the reaction.

Before proceeding, we hav to split the ionic compound; LiCl into it's component ions. So we have;

Cl₂ + 2Li ⟶ 2Li⁺Cl⁻

This leaves us with;

Cl₂ ⟶ 2Cl⁻ ............................... i

2Li ⟶ 2Li⁺ .............................. ii

Oxidation reactions can be identified by the increase in oxidation number and decrease in the case of reduction.

in reaction i, there is a decrease in oxidation number from 0 to -1. This is the reduction half equation,

in reaction ii, there is an increase in oxidation number from 0 to +1. This is the oxidation half equation

In terms of electrons, we have to even the charge;

Oxidation = Loss of electrons

Reduction = Gain of electrons

The half equations are given as;

Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation

2Li + 2e⁻ ⟶ 2Li⁺ Reduction

6 0

3 years ago

Radiant heat transfers energy

il63 [147K]

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3 years ago

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What is inside the nucleus of an atom

Tamiku [17]

The nucleus of a atom contains the protons and the neutrons all most all of the atoms mass is weigh by the protons and the neurons.

3 0

4 years ago

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