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xeze [42]
2 years ago
7

1. How much NaCl would you need to weigh out to prepare 100 mL of a 2.5% NaCl solution?

Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
7 0
<h3><u>Answer</u> - 5.85g</h3>

<h3><u>Solution</u> - </h3>

Using the concentration equation, m = CVMo where m is the mass of solute, C is the concentration (M), V is the volume of solution (L) and Mo is molar mass of the solute (g):

Given,

Molar mass of NaCl, M = 58.5g/mol

Volume of the solution, V= 100ml = 0.1L

Molarity, S= 1M

Mass of the solute,W=?

W=SMV

= 1×58.5×0.1

= 5.85g

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Explanation:

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3 years ago
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
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Answer:

49^oC

Explanation:

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

\Delta H^o_{vap} = 43988 J/mol\cdot \frac{1 mol}{18.016 g} = 2441.6 J/g

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

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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
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D. Producers im sure of it hope im right

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As the temperature of a liquid goes up, the average speed of its particles .This means that the kinetic energy of the particles
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The kinectic energy would be decreasing.
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In the molecule ClF3, chlorine makes three covalent bonds. Therefore, three of its seven valence electrons need to be unpaired.
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Answer:

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