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WINSTONCH [101]

2 years ago

13

If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, what will

the molarity of the solution be?

2 answers:

tatiyna2 years ago

8 0

Answer:

Explanation:

Molarity = number of moles / volume

If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, which is half of 550 mL, the molarity of the solution with the same number of moles of KCl is 3.5 * 2 = 7.00 M

Likurg_2 [28]2 years ago

4 0

C1=3.5M
C2=?
V1=550mL
V2=275mL

$\\ \rm\rightarrowtail C1V1=C2V2$

$\\ \rm\rightarrowtail 3.5(550)=275C2$

$\\ \rm\rightarrowtail C2=3.5(2)$

$\\ \rm\rightarrowtail C2=7M$

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Suppose that 0.48 g of water at 25 ∘ C condenses on the surface of a 55- g block of aluminum that is initially at 25 ∘ C . If th

mamaluj [8]

Answer:

$49^oC$

Explanation:

At $25^oC$ , the heat of vaporization of water is given by:

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The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:

$Q_1 = \Delta H^o_{vap} m_w$

The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:

$Q_2 = c_{Al}m_{Al}(t_f - t_i)$

According to the law of energy conservation, the heat lost is equal to the heat gained:

$Q_1 = Q_2$ or:

$\Delta H^o_{vap} m_w = c_{Al}m_{Al}(t_f - t_i)$

Rearrange for the final temperature:

$\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i$

We obtain:

$\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i = c_{Al}m_{Al}t_f$

Then:

$t_f = \frac{\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i}{c_{Al}m_{Al}} = \frac{2441.6 J/g\cdot 0.48 g + 0.903 \frac{J}{g^oC}\cdot 55 g\cdot 25^oC}{0.903 \frac{J}{g^oC}\cdot 55 g} = 49^oC$

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3 years ago

What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water?

dusya [7]

Answer:

The concentration of KBr is $C = 0.07899 \ mol L^{-1}$

Explanation:

From the question we are told that

The mass of KBr is $m_{KBr} = 2.35 \ g$

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This implies that the volume of the solution is $V = 250 mL$

The number of moles of KBr is

$n = \frac{m_{KBr}}{M_{KBr}}$

Substituting values

$n = \frac{2.35}{119}$

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The concentration of KBr is mathematically represented as

$C = \frac{0.01975}{0.250}$

$C = 0.07899 \ mol L^{-1}$

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3 years ago

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Katarina [22]

Answer:

The answer to your question is given below.

Explanation:

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