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SOVA2 [1]
3 years ago
8

Which factor below does not affect how fast a solute dissolves in a solvent?

Chemistry
2 answers:
Anit [1.1K]3 years ago
3 0
Stirring this is because the three elements are factors affecting dissolving of a solvent. Eg temprature affects in hotness or coldness, Particle size affects whether it is big or small while quantity of soluble affects by the amount
son4ous [18]3 years ago
3 0

Answer: The answer is B. Particle size of the solvent

Explanation: Solvent is usually in a liquid state (or sometimes in gaseous or solid-state). It is used in dissolving solutes (which are usually in solid-state). Although, the solute can also be in the form of a liquid and gaseous state.

When a solute dissolves in a solvent, a solution is formed. An example is when you have a salt (solute) dissolved in water(solvent) it will form a saline solution( which of course is the solution).

Factors that determine the rate a solute dissolves in a solvent include; temperature, stirring, the particle size of solute, the quantity of solute.

Temperature affects the rate of dissolving because, in a warmer solvent, solute has more energy to move compared in a cool solvent.

Stirring helps in the distribution of a solute in a solvent. When u agitate or stir, the solute will dissolve faster in the solvent.

The particle size of solute affects the rate of dissolving because a smaller particle size of a solute will dissolve faster than a solute with larger particle size.

Quantity of solute can also be a factor because when you have different quantities of solutes in a particular quantity of the same solvent, the rate of dissolving will be different provided they undergo the same factors I.e the solute with higher quantity will dissolve slower compared to the solute with less quantity provided the solvent is in the same temperature state.

Since solvents are usually in a liquid state, the particle size of the solvents won't be a factor of the rate at which the solutes will dissolve.

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A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg?
statuscvo [17]

Answer:

Final volume = V₂ =  8.2L

Explanation:

Given data:

Initial volume = 12.3 L

Initial pressure = 40.0 mmHg

Final volume = ?

Final pressure = 60.0 mmHg

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

40.0 mmHg × 12.3 L = 60.0 mmHg × V₂

V₂ = 40.0 mmHg × 12.3 L / 60.0 mmHg

V₂ =  492 mmHg. L /  60.0 mmHg

V₂ =  8.2L

7 0
3 years ago
Which elements are present in all organic compounds? 1. Hydrogen and oxygen 2. Nitrogen and Carbon 3. Nitrogen and oxygen 4. Hyd
cupoosta [38]

Answer:

my answer is num:3 Hydrogen and oxygen.

8 0
3 years ago
PLEASE HELP- I WILL GIVE BRAINLIEST AND THANKS
julsineya [31]

Answer:

The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams

Explanation:

The question is with regards to density calculations

The density of the given sodium carbonate solution, ρ = 0.4 g/dm³

The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³

Density \ of \ an \ object, \rho  = \dfrac{The \ mass \ of \ the \ object, \ m }{\ The \ volume \ of \ the \ object, \ V }

\rho = \dfrac{m}{V}

Therefore, we have;

The \ Density \ of \ the \ sodium \ carbonate, \ \rho  = 0.4 \ g/dm^3 =  \dfrac{m }{ 0.01 \ dm^3 }

The mass, "m", of the sodium carbonate in  = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g

The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.

8 0
3 years ago
Why do different elements have different colored flames when they are heated?
Mademuasel [1]
It depends on their energy levels.
5 0
3 years ago
For a reaction in a voltaic cell, both ΔH° and ΔS° are positive. Given the following equations for standard free energy, ΔG° =ΔH
nika2105 [10]

Answer:

The standard cell potential increases with increasing temperature.

Explanation:

Equatio 1: ΔG° =ΔH° -TΔS°

Equation 2:  ΔG° = -nFE°

Isolating E° in equation 2:

E° = - ΔG° / nF (Equation 3)

Substituting equation 1 in equation 3:

E° = (- ΔH° +TΔS°)/ nf

We can rearrange the equation:

E° = (ΔS°/nF) T + (ΔH°/nF)

Now it is clear that the higher the temperature, the higher the standard cell potential.

5 0
4 years ago
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