In what state of matter do molecules bounce off one another rapidly and freely

HELP ME PLEASE ASAP. I'LL MARK YOU BRAINLIEST <br> ITS NUMBER 23.

Ivanshal [37]

Answer:

The answer is A

Explanation:

5 0

3 years ago

1. Find the masses of the following amounts.

In-s [12.5K]

The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

A mole is a very important unit of measurement that chemists use.

A mole of something means you have 6.023 x 10 ²³ of that thing.

For 2.15 mol of hydrogen sulphide (H₂S) :

1 mole hydrogen sulphide (H₂S) = 34.08088 grams

Therefore,

2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol

= 73.272 gm

For 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) ;

1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams

Therefore,

3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol

= 1.82 gm

Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

Learn more about mole here ;

brainly.com/question/21323029

#SPJ1

7 0

2 years ago

1) A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10.000 g was heated until it reached a constant

vodomira [7]

The masses of the components are obtained as;

Sodium hydrogen carbonate = 3.51 g
Sodium carbonate = 8.708 g

<h3>What is decomposition?</h3>

The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.

Now putting down the equation of the reaction, we have;

$2NaHCO_{3} (s) ----- > Na_{2} CO_{3} (s) + CO_{2} (g) + H_{2} O(g)$

Now, the loss in mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g - 8.708 g = 1.292 g

Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol

= 3.51 g

Learn more about anhydrous sodium carbonate :brainly.com/question/20479996

#SPJ1

6 0

1 year ago

How do you know if a reaction is endothermic

Karo-lina-s [1.5K]

If the temperature rises in a reaction. Exothermic is if it loses heat.

Have a nice day, brainliest would be fantastic.

7 0

3 years ago

Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.5

e-lub [12.9K]

Answer:

Buffer B has the highest buffer capacity.

Buffer C has the lowest buffer capacity.

Explanation:

An effective weak acid-conjugate base buffer should have pH equal to $pK_{a}$ of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.

Acetic acid is a weak acid and $CH_{3}COO^{-}$ is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):

$pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

$pK_{a}(CH_{3}COOH)=4.74$

Buffer A: $pH=4.74+log(\frac{0.10}{0.10})=4.74$

Buffer B: $pH=4.74+log(\frac{0.30}{0.30})=4.74$

Buffer C: $pH=4.74+log(\frac{0.10}{0.50})=4.04$

So, both buffer A and buffer B has same pH value which is also equal to $pK_{a}$ . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.

The pH of buffer C is far away from $pK_{a}$ . Therefore, buffer C has the lowest buffer capacity.

6 0

3 years ago