All categories

3 years ago

As the tempreture of a liquid increases the solubility of a liquid of that liquid

1 answer:

8 0

As the temperature increases, the solubility of the solute in the liquid also increases. This is due to the fact that the increase in energy allows the liquid to more effectively break up the solute. The additoin of energy also shifts the equilibrium of the reation to the right since it takes energy to dissolve most things and you are adding more of it (this is explained with Le Chatlier principles).

I hope this helps and also I assumed that your question involved the solubility of an ionic substance in a solvent like water. If that was not your question feel free to say so in the comments so that I can answer your actually question.

You might be interested in

How many moles of copper are present in a sample of tennantite with a mass of 2290 grams?

grin007 [14]

Answer:

18.58 moles of Cu

Solution:

Data Given:

Chemical Formula of Tennantite = Cu₁₂As₄S₁₃

Mass of Cu₁₂As₄S₁₃ = 2290 g

M.Mass of Cu₁₂As₄S₁₃ = 1479.06 g.mol⁻¹

Step 1: Calculate Moles of Cu₁₂As₄S₁₃ as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 2290 g ÷ 1479.06 g.mol⁻¹

Moles = 1.548 mol

Step 2: Calculate Moles of Cu,

As,

1 mole of Cu₁₂As₄S₁₃ contains = 12 moles of Cu

So,

1.548 mol of Cu₁₂As₄S₁₃ will contain = X moles of Cu

Solving for X,

X = (1.548 mol × 12 mol) ÷ 1 mol

X = 18.58 moles of Cu

5 0

3 years ago

Calculate the percent ionization of a 0.15 M benzoic acid solution in pure water and in a solution containing 0.10 M sodium benz

Hoochie [10]

Answer:

% ionization for benzoic acid = 0.08%

% ionization for sodium benzoate = 2.5%

The percentage ionization differ significantly because benzoic acid is a weak acid while sodium benzoate is a salt of benzoic acid. Their extent of dissociation also differ because they were compared in different solutions

Explanation:

Ka for pure water = 1.0 * 10-⁷

Ka for sodium benzoate = 6.5*10-⁵

1. For benzoic acid (C6H5COOH)

C6H5COOH ==== C6H5COO‐ + H+

0.15M 0 0

0.15-x x x

Ka = [C6H5COO-] [H+] / [C6H5COOH]

Ka = [X] [X] / 0.15 - X

1.0*10-⁷ = [X]² / 0.15 - x

But x is negligible compared to 0.15,

(1.0*10-⁷)*0.15 = x²

Take square root of both sides,

X = 1.22 * 10-⁴

% ionization = ( [H+] / [C6H5COOH] ) * 100

% ionization = (1.22*10-⁷ / 0.15) * 100

% ionization = 0.08%

2. For C6H5COONa

Note: I will not repeat the same procedure of dissociation again since they're basically the same just the difference in ions

Ka for C6H5COONa = 6.5*10-⁵

6.5*10-⁵ = [X]² / (0.10 - X)

Cross-multiply both sides;

(6.5*10-⁵ * 0.10) = X²

Take square root of both side,

X= 2.5*10-³

% ionization = (2.5*10-³ / 0.10) *100

% ionization = 2.5%

5 0

3 years ago

Which of the following statements provide evidence to support the big bang theory? Select the two correct answers. i need help p

Leona [35]

Answer:

See explaination

Explanation:

CBR (Cosmic Background Radiation)-

Uniform energy that is heard everywhere in the universe. It supports the Big Bang Theory because we can assume that it also started in the same place.

Hubble's Law-

Hubble's Law states that the planets are getting farther away from eachother, therefore, the universe is expanding and must have started as a smaller piece.

8 0

3 years ago

The following redox reaction is conducted with [Al3+] = 0.12 M and [Mn2+] = 1.5 M.

Ronch [10]

Answer:

0.47V

Explanation:

2 Al(s) + 3 Mn2+(aq) → 2 Al3+(aq) + 3 Mn(s)

n= 6 ( six moles of electrons were transferred)

Q= [Red]/[Ox] but [Red]= 1.5M, [Ox] = 0.12 M

Q= 1.5/0.12= 12.5

From Nernst equation:

E= E°cell- 0.0592/n log Q

E°cell= 0.48 V

E= 0.48 - 0.0592/6 log (12.5)

E= 0.47V

6 0

4 years ago

The yearly amounts of carbon emissions from cars in Belgium are normally distributed with a mean of 13.9 gigagrams per year and

zavuch27 [327]

Answer:

The correct option is;

c. 0.167

Explanation:

The parameters given are;

The mean, μ = 13.9 Gigagrams/year

The standard deviation, σ = 5.8 Gigagrams/year

The z-score formula is given as follows;

$z = \dfrac{x - \mu }{\sigma }$

Where:

x = Observed score = 11.5 Gigagrams/year

We have;

$z = \dfrac{11.5 - 13.9 }{5.8 } = \dfrac{-2.4 }{5.8 } = 0.4138$

From the z-score table relations/computation, the probability (p-value) = 0.6605

Where:

x = Observed score = 14 Gigagrams/year

We have;

$z = \dfrac{14- 13.9 }{5.8 } = \dfrac{0.1}{5.8 } = 0.01724$

From the z-score table relations/computation, the probability (p-value) = 0.4931

Therefore, the probability, $p_{ca}$ , that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 15.5 Gigagrams/year and 14 Gigagrams/year = The area under the normal curve bounded by the p-values for the two amounts of carbon emission

Which gives;

$p_{ca}$ = 0.6605 - 0.4931 = 0.1674 ≈ 0.167

Therefore, the correct option is c. 0.167.

4 0

4 years ago