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3 years ago

Balanceo de k20 + H3BO3

Chemistry

1 answer:

kipiarov [429]3 years ago

5 0

Whats your question? id love to help but I’m not sure I understand

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Which is an example of flow of heat through conduction

JulsSmile [24]

Answer:

a metal spoon left in boiling water

Explanation:

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3 years ago

What is the molality of a solution composed of 342 moles of sugar (C12H22O11) dissolved in 171 kilograms of water (H20)?

4vir4ik [10]

Molality=mol/kg

342/171

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2 years ago

Honeybee venom is a solution that contains formic acid, which gives the venom an approximate ph of 5.0. which substance is more

Genrish500 [490]

Those substances which has pH less than 5 are more acidic than Honeybee Venom. Some of them are listed below along with their pH scale,

1) <span>Acid rain and Tomato Juice has a pH of 4.

2) Soda and Orange Juice has a pH of 3.

3) Vinegar and Lemon Juice has pH of 2.

4) Gastric acid has a pH of 1.

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4 years ago

What is the pH of a solution in which [H 3O +] = 3.8 × 10 -8 M?

7nadin3 [17]

Answer:

<h3>The answer is 7.42 </h3>

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ { H_3O}^{+}]$

From the question we have

$pH = - log(3.8 \times {10}^{ - 8} ) \\ = 7.420216...$

We have the final answer as

Hope this helps you

4 0

3 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

3 years ago

Balanceo de k20 + H3BO3​

Balanceo de k20 + H3BO3