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jekas [21]
2 years ago
5

What is the concentration of a solution that is 0.25 moles of HCL added to 3.00L of water?

Chemistry
1 answer:
Lina20 [59]2 years ago
3 0

Answer: 0.08 M

Explanation:

\text{molarity}=\frac{\text{moles of solute}}{\text{liters of solution}}\\\\\text{molarity}=\frac{0.25}{3.00}\\\\\text{molarity}=\boxed{0.08 \text{ M}}

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Respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce use able cellular energy. This energy is used to fuel growth and all of the normal cellular functions.

Short Answer

They combine to produce cellular energy.

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Mechanically weathered rock is broken down into smaller _________________________. (four word phrase)
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3 years ago
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The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react
Kazeer [188]

Answer:

The percent yield of  chloro-ethane in the reaction is 82.98%.

Explanation:

C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl

Moles of ethane = \frac{300.0 g}{30 g/mol}=10 mol

Moles of chlorine gases =\frac{650.0 g}{71 .0 g/mol}=9.1549 mol

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

\frac{1}{1}\times 9.1549 mol=9.1549 mol of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of  chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%

The percent yield of  chloro-ethane in the reaction is 82.98%.

6 0
3 years ago
Why are halogens reactive?
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