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Vsevolod [243]
3 years ago
7

Sasha did an experiment to study the solubility of two substances. She poured 100 mL of water at 20 °C into each of two beakers

labeled A and B. She put 50 g of Substance A in the beaker labeled A and 50 g of Substance B in the beaker labeled B. The solution in both beakers was stirred for 1 minute. The amount of substance left undissolved in the beakers was weighed. The experiment was repeated for different temperatures of water and the observations were recorded as shown.
Part 1: Which substance has a higher solubility?

Part 2: Explain your answer for Part 1.
Chemistry
1 answer:
amm18123 years ago
8 0

Answer:

A

Explanation:

The solubility of a substance is directly proportional to the temperature. It means that solubility will increase with the increases in temperature. At higher temperature, the kinetic energy increased that allow the solvent molecules to break the solute particles more effectively.

Substance A has a higher solubility because the weight of substance A measured at the end of the experiment is less than the weight of substance B.

Hence, the correct answer is A.

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Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o
sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4}  } = 0.0106 M

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%

4 0
3 years ago
What pressure in atmospheres(atm) is equal to 45.6 kPa?
postnew [5]
1 kpa = 0.0098692327 atm so just multiply that by 45.6
4 0
4 years ago
69. Sequencing Arrange these hydrates in order of
Kobotan [32]

In order of  increasing percent water content:CoCl₂.6H₂O, Ba(OH)₂.8H₂O, MgSO₄.7H₂O

<h3>Further explanation</h3>

\tt \%element=\dfrac{Ar~element}{MW~compound}\times 100\%

CoCl₂.6H₂O.MW=237.90 g/mol

6H₂O MW = 6.18=108 g/mol

\tt \%H_2O=\dfrac{108}{237.9}\times 100\%=45.4\%

MgSO₄.7H₂O.MW=246.48 g/mol

MW 7H₂O = 7.18=126 g/mol

\tt \%H_2O=\dfrac{126}{246.48}\times 100\%=51.1\%

Ba(OH)₂.8H₂O MW=315.48 g/mol

MW 8H₂O = 8.18=144 g/mol

\tt \%H_2O=\dfrac{144}{315.48}\times 100\%=45.6\%

4 0
3 years ago
When 2.5 mol of O2 are consumed in their reaction, ________ mol of CO2 are produced
Vika [28.1K]

The given question is incomplete. The complete question is:

The combustion of propane (C3H8) in the presence of excess oxygen yields CO_2 and H_2O

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

When only 2.5 mol of O_2 are consumed in order to complete the reaction, ________ mol of CO_2 are produced.

Answer: Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

Explanation:

The balanced chemical equation is:

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

According to stoichiometry :

5 moles of O_2 produce = 3 moles of  CO_2

Thus 2.5 moles of O_2 will produce = \frac{3}{5}\times 2.5=1.5 moles of  CO_2

Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

4 0
3 years ago
Which of these is a chemical property?
ankoles [38]

Answer:

The second choice, or flammability.

Explanation:

The flammability of something is how easy it is for it to burn or ignite.

7 0
3 years ago
Read 2 more answers
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