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weqwewe [10]
2 years ago
10

4. You have a very concentrated solution (12 M) of potassium chloride (KCI). You need it to be at the least concentration possib

le for the experiment you are
about to conduct. The problem is you forgot to order lab supplies, so you only have 2 L of distilled water left. What would be the final concentration if you
added the two 2 L of distilled water to the 0.5 L of 12 M KCI?
a 3.0 M
b 24 M
C 2.4M
d 48 M
Chemistry
1 answer:
Alenkinab [10]2 years ago
4 0

Answer:

C. 2.4 M

Explanation:

Because you started with 12M solution of KCl, that means that there was a concentration of 12 mols of KCl per Liter of solution. (12mol/1L =12 M). Since there was only 0.5L of solution, there was only 6 mols of KCl because there is only 12 mols per 1 Liter, so half of that volume would have half the amount of solute to keep that true. (6mol/.5L = 12 M). With the new water added, the formula becomes (6mols KCl/ 2.5L of water =2.4 M)

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
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Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

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What does "hidden phenomena" mean in science?
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Answer:

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Propylethylene would be the answer

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For a reaction at equilibrium, which change can increase the rates of the forward and reverse reactions?a decrease in the concen
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Question requires a change resulting in an increase in both forward and reverse reactions. Now lets discuss options one by one and see there impact on rate of reactions.

1) <span>A decrease in the concentration of the reactants:
                                                                                       
When concentration of reactant is decreased it will shift the equilibrium in Backward direction, so resulting in increasing the backward reaction and decreasing the forward direction. Hence, this option is incorrect.

2) </span><span>A decrease in the surface area of the products:
                                                                               Greater the surface Area greater is the chances of collision and greater will be the rate of reaction. As the surface area of products is decreased it will not favor the backward reaction. Hence again this statement is incorrect according to given statement.

3) </span><span>An increase in the temperature of the system:
                                                                             An increase in temperature will shift the reaction in endothermic side. Hence, if the reaction is endothermic, an increase in temperature will increase the rate of forward direction or if the reaction is exothermic it will increase the rate of reverse direction. Hence, this option is correct according to given statement.

4) </span><span>An increase in the activation energy of the forward reaction:
                                                                                                   An increase in Activation energy will decrease the rate of reaction, either it is forward or reverse. So this is incorrect.

Result:
          Hence, the correct answer is,"</span>An increase in the temperature of the system".
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2 years ago
How many moles of water h2o are present in 75.0 g h2o?
nikklg [1K]
4.17 moles. Good luck! :)
7 0
3 years ago
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