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timofeeve [1]
3 years ago
14

If you add meoh to cocl2(alc)2, the cocl2(alc)2 concentration will change. what are the two reasons for this concentration chang

e?
Chemistry
1 answer:
Vaselesa [24]3 years ago
5 0
There are two reasons that account for the change in concentration. One is by the physical change and second is by the chemical change. The physical change refers to the mixture of the two compounds. The concentration is the amount of that certain substance to the whole mixture. Since the mixture increases because you add them up, the concentration decreases. Second, these compounds react with each other once they interact. Hence, their amounts would be consumed in order to produce the products.
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(b) What is the pOH of 1.35x10⁻³ M HCl? Is the solution neutral, acidic, or basic?
nika2105 [10]

The pOH scale is similar to the pH scale in that a pOH of 7 is indicative of a neutral solution. A basic solution has a pOH less than 7, while an acidic solution has a pOH of greater than 7

<h3>What is pOH scale?</h3>

Based on the relative concentration of hydrogen and hydroxyl ions, the pH and pOH scales are used primarily in chemistry, biology, and soil science to describe and measure the acidity and basicity of neutral, acidic, and basic solutions in water.

The terms pH and pOH scale are frequently used and crucial for determining the chemical, biological, and soil health by converting the values of hydrogen or hydroxyl ion concentrations or by utilizing pH paper and meters.

The pH scale or chart, which measures the concentration of hydrogen and hydroxyl ions in acidic and alkaline solutions, was initially developed by Sorensen.

To learn more about pOH scale from the given link:

brainly.com/question/16631843

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5 0
2 years ago
A sample of methane collected when the temp was 30 C mmHg measures 398 mL. What would be the volume of the sample at -5 C and 61
Levart [38]
<h2>Question </h2>

A sample of methane collected when the temp was 30 C and 760mmHg measures 398 mL. What would be the volume of the sample at -5 C and 616 mmHg pressure

<h2>Answer:</h2>

434.32mL

<h2>Explanation:</h2>

Using the combined gas law:

\frac{PV}{T} = k

Where;

P = Pressure

V = Volume

T = Temperature

k = constant.

It can be deduced that:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = k           ---------------------(i)

Where:

P₁ and P₂ are the initial and final pressures of the given gas

V₁ and V₂ are the initial and final volumes of the given gas

T₁ and T₂ are the initial and final temperatures of the gas.

<em>From the question:</em>

the gas is methane

P₁  = 760mmHg

P₂ = 616mmHg

V₁ = 398mL

V₂ = ?

T₁ = 30°C = (30 +273)K = 303K

T₂ = -5°C = (-5 +273)K = 268K

Substitute these values into equation (i) as follows;

\frac{760*398}{303} = \frac{616*V_2}{268}

Solve for V₂

V₂ = \frac{760*398*268}{616*303}

V₂ = 434.32mL

Therefore, the volume of the sample at -5C and 616mmHg pressure is 434.32mL

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