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3 years ago

The normal boiling point for h2se is higher than the normal boiling point for h2s . this can be explained by

1 answer:

8 0

H2se is a ionic bond which requires higher energy to change into a gas, while h2s is a covalent bond requiring less energy

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The industrial preparation of ozone (03) uses oxygen gas (O2), as shown in the equation below. 302(8) electricity 203(g) A If 96

Paul [167]

The percent yield of the industrial preparation of ozone is the 12%.

<h3>How do we calculate the percent yield?</h3>

Percent yield of any reaction will be calculated by using the below equation:

% yield = (Actual yield / Theoretical yield) × 100%

Given chemical reaction is:

3O₂(g) → 2O₃(g)

Moles of oxygen will be calculated as:

n = W/M, where

W = given mass = 96kg = 96,000g
M = molar mass = 32g/mol

n = 96,000/32 = 3000mol

From the stoichiometry of the reaction it is clear that:

3 moles of oxygen = produces 2 moles of ozone

3000 moles of oxygen = produces 2/3(3000)=2000 moles of ozone

Mass of 2000 moles of ozone = (2000mol)(48g/mol) = 96,000g = 96kg

Actual yield of ozone = 11.5kg (given)

Now percent yield will be calculated as:

% yield = (11.5 / 96) × 100% = 11.97% = 12%

Hence required percent yield is 12%.

To know more about percent yield, visit the below link:

brainly.com/question/20349874

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5 0

2 years ago

How much 10.0 M HNO must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH

Naya [18.7K]

Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

Moles of acetic acid = 0.0100 M × 1.00 L

= 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

= 0.100 mol

When $HNO_{3}$ is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles $HNO_{3}$ )

= 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles $HNO_{3}$ )

= 0.100 mol - x

According to Henderson - Hasselbalch equation,

pH = $pK_{a} + log\frac{[conjugate base]}{[weak acid]}$

pH = $pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}$

4.95 = 4.75 + $log\frac{(0.100 mol - x)}{(0.0100 mol + x)}$

$log\frac{(0.100 mol - x)}{(0.0100 mol + x)}$ = 4.95 - 4.75

= 0.20

$\frac{(0.100 mol - x)}{(0.0100 mol + x)}$ = antilog (0.20)

= 1.6

Hence, x = 0.032555 mol

Therefore, moles of $HNO_{3}$ = 0.032555 mol

volume of $HNO_{3}$ = $\frac{moles HNO_{3}}{molarity of HNO_{3}}$

= $\frac{0.032555 mol}{10.0 M}$

= 0.0032555 L

or, = 3.25 (as 1 L = 1000 mL)

Thus, we can conclude that volume of $HNO_{3}$ added is 3.26 mL.

5 0

3 years ago

Please help! Thank you!

MArishka [77]

It has to be oxygen as it has a charge of -2 and magnesium has a charge of +2 so when they bond it would be MgO

5 0

3 years ago

598mm to m<br><br>express your answer to three significant figures and include the appropriate units

nasty-shy [4]

0.598 m

598 mm × (1 m/1000 mm) = 0.598 m

8 0

4 years ago

Please help! Will mark brainliest!

seraphim [82]

I believe its A. please correct me if im wrong! <3

3 0

3 years ago

Read 2 more answers