How are electrons arranged around an atom?

When a 0. 1 m aqueous solution of hydrocyanic acid, hcn, reaches equilibrium, the ph is measured to be 5. 20. using this informa

Sav [38]

When a 0. 1 m aqueous solution of hydrocyanic acid, HCN, reaches equilibrium, the ka for hydrocyanic acid is 3.969 x 10⁻¹⁰.

<h3>What is ka value?</h3>

It's the value of equilibrium constant for the dissociation of ions into a solution. The more the Ka value the more will be dissociation.

Ka = [H₃O⁺]² / [HCN] [H₃O⁺]

The pH is 5.20

-log [H₃O⁺] = 5.20

Putting antitlog both side.

The value will be 6.30 x 10⁻⁶

Ka = (6.30 x 10⁻⁶)² / 0.1 - 6.30 x 10⁻⁶

0.1 - 6.30 x 10⁻⁶ = 0.1

Ka = 3.969 x 10⁻¹⁰

Thus, the Ka value for hydrocyanic acid is 3.969 x 10⁻¹⁰.

To learn more about ka value, refer to the link:

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7 0

1 year ago

How many atoms are in 766.3 grams of Li?

34kurt

1 mol of any substance is made of 6.022 x 10²³ units, these units could be atoms making up elements or molecules making up a compound.

In this case. 1 mol of Li is made of 6.022 x 10²³ atoms of Li

The molar mass of Li is 6.94 g/mol

Therefore mass of 1 mol of Li is 6.94 g

In 6.94 g of Li there are - 6.022 x 10²³ atoms

Then in 766.3 g of Li there are - 6.022 x 10²³ atoms / 6.94 g x 766.3 g = 665 x 10²³

There are 6.65 x 10^25 atoms of Li

6 0

3 years ago

A 1.525g sample of a compound between nitrogen and hydrogen contains 1.333 g of nitrogen. Calculate its empirical formula. The e

vredina [299]

Answer:

NH2.

Explanation:

The mass of hydrogen in the sample = 1.525 - 1.333 = 0.192g.

Dividing the 2 masses by the relative atomic mass of hydrogen and nitrogen:

H: 0.192 / 1.008 = 0.1905

N: 1.333 / 14.007 = 0.09517

The ratio of N to H = 0.09517 : 0.1905

= 1 : 2.

So the empirical formula is NH2.

5 0

3 years ago

What is a liquid? can a liquid change shape or volume

NeX [460]

Yes a liquid can change into a volume based on the amount of whatever liquid

7 0

3 years ago

Read 2 more answers

Calculate the pH of adding 20 mL of 1 M NaOH solution to 100 mL of a 1 M acetic acid (CH3COOH) solution and 880 mL of distilled

Dafna1 [17]

Answer: The pH of the solution is 4.14

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For NaOH:

Molarity of NaOH = 1 M

Volume of solution = 20 mL

Putting values in above equation, we get:

$1M=\frac{\text{Moles of NaOH}\times 1000}{20mL}\\\\\text{Moles of NaOH}=\frac{1\times 20}{1000}=0.02mol$

For acetic acid:

Molarity of acetic acid solution = 0.26 M

Volume of solution = 100 mL

Putting values in above equation, we get:

$1M=\frac{\text{Moles of acetic acid}\times 1000}{100mL}\\\\\text{Moles of acetic acid}=\frac{1\times 100}{1000}=0.1mol$

The chemical reaction for NaOH and acetic acid follows the equation:

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

Initial: 0.100 0.020

Final: 0.080 - 0.020

Volume of solution = 20 + 100 = 120 mL = 0.120 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=\frac{0.020}{0.120}$

$[CH_3COOH]=\frac{0.080}{0.120}$

pH = ?

Putting values in above equation, we get:

$pH=4.74+\log(\frac{0.020/0.120}{0.080/0.120})\\\\pH=4.14$

Hence, the pH of the solution is 4.14

5 0

3 years ago