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4 years ago

The concentration of the solid phase _____ the quantity present.

2 answers:

5 0

The concentration of the solid phase always remains the same regardless of the quantity present. Most of the chemical reactions and separation systems involve the transport of more or one components in solid phase.
For example, in the process of solid catalyzed reactions, the result of solid phase transport might be negligible though it is hard to identify it in the rate of data in the analysis.

Stels [109]4 years ago

4 0

Answer: Option (b) is the correct answer.

Explanation:

A solution that contains more number of solute particles than the solvent particles is known as a concentrated solution.

For example, concentrated solution of sulfuric acid.

On the other hand, a solution that contains less number of solute particles than the solvent particles is known as a dilute solution.

For example, a dilute solution of HCl.

Hence, we can conclude that concentration of a substance changes with change in the amount of solute.

Therefore, the concentration of the solid phase changes with the amount of the quantity present.

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3 years ago

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

Answer: The mass of glucose in final solution is 1.085 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

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Solve for V2 when P1 = 1.0 atm V1 = 22.4 L and P2 = 4.0 atm This is a Boyle’s law gas problem

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P1/V1=P2/V2
1/22.4=4/x
X=4 multiple by 22.4
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