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3 years ago

Explain the electrolysis of an aqueous solution. (What is produced at each electrode and why its produced) Thank you

Chemistry

1 answer:

muminat3 years ago

7 0

Answer:

Oxidation at the anode. Lost of electrons. Oxygen loses electrons.

Reduction at the cathode. Gain electrons. Hydrogen gains electrons.

Explanation:

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A sample of gas has an initial volume of 3.62 L at a pressure of 0.987 atm. If the volume of the gas decreases to 2.50 L, what w

statuscvo [17]

Answer:

0.033 atm

Explanation:

PV=nRT

rule of 1 for T and n

P= nRT/V

P= 1(.08206)1)/ 2.50

0.033 atm = pressure

3 0

3 years ago

What is the pH of a 0.640 M solution of C₅H₅NHBr (Kb of C₅H₅N is 1.7 × 10⁻⁹)?

Elden [556K]

The pH of a 0.64 M solution of pyridine (C₅H₅N) is 9.52.

A figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values more alkaline.

The equation for the protonation of the base pyridine is the following:

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻ (1)

Kb = 1.7 × 10⁻⁹ (Given)

To calculate the pH of the solution we need to use the following equation:

pH + pOH = 14

=14 - [-log[OH⁻]]

= 14 + log[OH⁻]

Now, we need to find the concentration of the OH⁻ ions. Since pyridine is a weak base, at the equilibrium we have (eq 1):

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻

0.64 - x x x

After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

1.7 × 10⁻⁹ =[C₅H₅NH⁺] [OH⁻] / [C₅H₅N]

= x . x / 0.64-x

1.7 × 10⁻⁹ (0.64-x) - x² = 0

Solving the above quadratic equation for x, we have :

x₁ = -3.32 x 10⁻⁵
x₂ = 3.32 x 10⁻⁵

Now, We can calculate the pH, after taking the positive value, x₂, (concentrations cannot be negative) and entering into above equation :

<em />

<em>pH = </em>14 + log[OH⁻]

= 14 + log (3.32 x 10⁻⁵)

= 9.52

Therefore, the pH of the solution of pyridine is 9.52.

Find more about pH here:

brainly.com/question/8834103?referrer=searchResults

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3 0

2 years ago

Consider the chemical equations shown here.

vichka [17]

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Explanation:

The overall equation for the reaction that produces P₄0₁₀ is :

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Now let us derive this equation:

Given equations:

P₄ $_{s}$ + 30₂ $_{g}$ ⇒ P₄0₆ $_{s}$ equation 1;

P₄ $_{s}$ + 50₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$ equation 2;

To get the overall combined equation, the equation 1 must be reversed and added to equation 2:

P₄0₆ $_{s}$ ⇒ P₄ $_{s}$ + 30₂ $_{g}$ equation 3

equation 2:

P₄ $_{s}$ + 50₂ $_{g}$ + P₄0₆ $_{s}$ ⇒ P₄0₁₀ $_{s}$ + P₄ $_{s}$ + 30₂ $_{g}$

cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

learn more:

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5 0

3 years ago

How many molecules are in 3.6 grams of NaCl?

raketka [301]

Answer:

$\boxed{\text{None}}$

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

$\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}$

Step 2. Convert moles to formula units

$\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}$

There are $\boxed{3.4 \times 10^{22} \text{ FU NaCl}}$ in 3.6 g of NaCl.

6 0

3 years ago

What is the empirical formula for a compound that has 57.5%

Rainbow [258]

Answer:

Explanation:

use %composition given to calculate emperical 4mular

8 0

3 years ago

Explain the electrolysis of an aqueous solution. (What is produced at each electrode and why its produced) Thank you ​

Explain the electrolysis of an aqueous solution. (What is produced at each electrode and why its produced) Thank you