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4 years ago

Particle with more energy move _________ than particles with less energy.

Chemistry

2 answers:

s2008m [1.1K]4 years ago

5 0

Faster and farther apart

dlinn [17]4 years ago

4 0

Its correct I took the test and the answer was faster and farther apart

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Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significa

Black_prince [1.1K]

Answer: The equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

Explanation:

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid $(H_2C_2O_4)$ is a weak acid and will dissociate 2 hydrogen ions.

The chemical equation for the first dissociation of oxalic acid follows:

$H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)$

Initial: 0.20

At eqllm: 0.20-x x x

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}$

We know that:

$Ka_1\text{ for }H_2C_2O_4=0.059$

Putting values in above equation, we get:

$0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

The chemical equation for the second dissociation of oxalic acid:

$HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)$

Initial: 0.083

At eqllm: 0.083-y 0.083+y y

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}$

We know that:

$Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}$

Putting values in above equation, we get:

$6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

7 0

3 years ago

Name of Metallic compound VSe

Komok [63]

Answer: vanadium selenide

Explanation:

5 0

3 years ago

Read 2 more answers

Complete the chemical equation related to the formation of sodium chloride

son4ous [18]

Answer:

sodium + chlorine --> sodium chloride

3 0

3 years ago

Calculate the number of moles of Na+ ions in 25g of sodium carbonate?

valkas [14]

Explanation:

First, we need to calculate the number of moles of sodium carbonate we have in a 25 g sample. To calculate this, we will

find the molar mass of sodium carbonate (Na2CO3):

⇒ 2 × Molar mass of sodium + Molar mass of carbon + 3×molar mass of oxygen

⇒ 2 × 23 + 12 + 3 × 16

⇒ 46 + 12 + 48

⇒ 106g/mol

Thus, the molar mass of Na2CO3 is 106g/mol.

Therefore, number of moles = 25 ÷ 106

=> 0.2358 mol

Now, we know that every mole of Na2CO3 have 0.2358 moles of Na+ ions. Hence, total moles of Na2CO3 is 0.4716 moles

Number of ions present = 6.022 × 1023 × 0.4716 mol = 2.84 × 1023ions

6 0

3 years ago

25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311

LuckyWell [14K]

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

5 0

4 years ago