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Shkiper50 [21]
3 years ago
10

Which of the following would most likely cause a chemical reaction to occur at a slower rate?

Chemistry
2 answers:
Keith_Richards [23]3 years ago
7 0

Answer:

D

Explanation: When you chill a substance when it is mixed with another substance, ( chilled vinegar mixed with sodium bicarbonate will react slower for instance) it reacts slower do to the molecules slowing down when chilled.

Burka [1]3 years ago
7 0
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 />

\huge\boxed{Option D}

<h2>Solution:</h2>

a liquid mixture in which the the solute which is in a smaller proportion is uniformly distributed within the the solvent which is the major component is called a solution.

<h2>_____________________________________</h2><h2>Option A</h2><h3>Stirring the substance together vigorously:</h3>

It increases the rate of reaction,  because when we stir the solute in the solution, the mixed solute particles are distributed more further than when they are left stationary, making the environment (which surrounds the solute) less concentrated by the dissolved solute thus the rate increases and more solute is mixed.

<h2>Option B</h2><h3>Dissolving the reacting substance in water:</h3>

No, A reacting substances  in water are those which can react with water that are Ionic metal. Water is made of ions H^+ and OH^-. Substances(solute), they mix or dissolve more because Solutes ions interact with the ions of a water(solvent), thus breaking the bonds between the solute and then ions of solute form bond with the ions of water. Thus the rate increases in water.

<h2>Option C</h2><h3>Increasing the temperature increases the rate of reaction:</h3>

No, because by the increase in temperature, the kinetic energy of more  molecules of a substance increase which results in reaching the threshold energy(minimum energy required by the molecules to change into product) of more molecules. So less molecules are left with low energy.

<h2>Option D</h2><h3>Decreasing the temperature decrease the rate of reaction.</h3>

Yes, the rate of reaction decreases because  the molecules are slower and collide less. Less molecules make the effective collisions due to which the kinetic energy of molecules is less, so less molecules make up to the threshold energy. So decrease in temperature lowers the rate of the reaction.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

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Answer:

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                                            3s²   3p¹

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"An aqueous CaCl2 solution has a vapor pressure of 83.1mmHg at 50 ∘C. The vapor pressure of pure water at this temperature is 92
Lynna [10]

Answer : The the concentration of CaCl_2 in mass percent is, 41.18 %

Solution : Given,

Molar mass of water = 18 g/mole

Molar mass of CaCl_2 = 110.98 g/mole

First we have to calculate the mole fraction of solute.

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

\fac{p^o-p_s}{p^o}=X_B

where,

p^o = vapor pressure of the pure component (water) = 92.6 mmHg

p_s = vapor pressure of the solution = 83.1 mmHg

X_B = mole fraction of solute, (CaCl_2)

Now put all the given values in this formula, we get the mole fraction of solute.

\fac{92.6-83.1}{92.6}=X_B

X_B=0.102

Now we have to calculate the mole fraction of solvent (water).

As we know that,

X_A+X_B=1\\\\X_A=1-X_B\\\\X_A=1-0.102\\\\X_A=0.898

The number of moles of solute and solvent will be, 0.102 and 0.898 moles respectively.

Now we have to calculate the mass of solute, (CaCl_2) and solvent, (H_2O).

\text{Mass of }CaCl_2=\text{Moles of }CaCl_2\times \text{Molar mass of }CaCl_2

\text{Mass of }CaCl_2=(0.102mole)\times (110.98g/mole)=11.32g

\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O

\text{Mass of }H_2O=(0.898mole)\times (18g/mole)=16.164g

Mass of solution = Mass of solute + Mass of solvent

Mass of solution = 11.32 + 16.164 = 27.484 g

Now we have to calculate the mass percent of CaCl_2

Mass\%=\frac{\text{Mass of}CaCl_2}{\text{Mass of solution}}\times 100=\frac{11.32g}{27.484g}\times 100=41.18\%

Therefore, the the concentration of CaCl_2 in mass percent is, 41.18 %

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