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Alex787 [66]
2 years ago
5

What volume of 0.555M KNO3 solution would contain 12.5 g of solute

Chemistry
1 answer:
Lorico [155]2 years ago
3 0

The volume of 0.555M KNO3 solution would contain 12.5 g of solute iss 223 mL.

<h3>What is the relationship between mass of solute and concentration of solution?</h3>

The mass of solute in a given volume of solution is related by the formula below:

  • Molarity = mass/(molar mass * volume)

Therefore, volume of solution is given by:

Volume = Mass /molarity * molar mass

Molar mass of KNO₃ = 101 g/mol

Volume = 12.5/(0.555 * 101)

Volume = 0.223 L or 223 mL

In conclusion, the volume of the solution is obtained from the molarity of solution as well as mass and molar mass of solute.

Learn more about molarity and volume at: brainly.com/question/26873446
#SPJ1

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What factors affect the dynamic state of equilibrium in a chemical reaction and how?
yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

  • External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

  • \Delta G the standard Gibbs energy change of the reaction, and
  • T the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

  • Changes in concentration influence the number of particles per unit space.
  • Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

  • Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

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In a percentage composition investigation a compound was decomposed into its elements: 20.0 g of calcium, 6.0 g of carbon, and 2
inysia [295]

The percentage composition of this compound : 40%Ca, 12%C and 48%O

<h3>Further explanation</h3>

Given

20.0 g of calcium,

6.0 g of carbon,

and 24.0 g of oxygen.

Required

The percentage composition

Solution

Total mass of compound :

=mass calcium + mass carbon + mass oxygen

=20 g + 6 g + 24 g

=50 g

Percentage composition :

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\tt \dfrac{20}{50}\times 100\%=40\%

  • C-carbon

\tt \dfrac{6}{50}\times 100\%=12\%

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