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yKpoI14uk [10]
3 years ago
15

Calculate the freezing point of a solution 1.25 g benzene (C6H6) in 125 g of chloroform (CHCl3).

Chemistry
1 answer:
posledela3 years ago
3 0

Answer:

The freezing point for the solution is -64.09°C

Explanation:

This problem can be solved, by the freezing point depression. This colligative problem shows, that the freezing point of a solution is lower than the freezing point of pure solvent.

ΔT = Kf.  m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryscopic constant, for chloroform is 4.68

T°freezing pure solvent = -63.5°C

m is mol/kg of solvent → molality

Let's determine the moles of benzene

1.25 g / 78 g/mol = 0.0160 mol

Let's convert the mass of solvent to kg

125 g . 1kg / 1000 g = 0.125 kg

m = 0.0160 mol / 0.125 kg → 0.128 m

Let's go to the formula to replace the data

-63.5°C - T° freezing solution = 4.68 °C/m . 0.128 m

T° freezing solution = - (4.68 °C/m . 0.128 m + 63.5°C)

T° freezing solution = - 64.09°C

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Dvinal [7]

Answer : The volume of hydrogen gas at STP is 4550 L.

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 100.0 atm

P_2 = final pressure of gas at STP = 1 atm

V_1 = initial volume of gas = 50.0 L

V_2 = final volume of gas at STP = ?

T_1 = initial temperature of gas = 27.0^oC=273+27.0=300K

T_2 = final temperature of gas at STP = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{100.0atm\times 50.0L}{300K}=\frac{1atm\times V_2}{273K}

V_2=4550L

Therefore, the volume of hydrogen gas at STP is 4550 L.

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3 years ago
How does a greenhouse keep radiant energy from escaping?
Rina8888 [55]
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The green house glass are insulated and they trap the infrared emitted by the objects inside the green house from escaping outside. Since the infrared have longer wave lengths, it is released slowly. 

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3 years ago
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5 0
3 years ago
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In what state of matter do molecules bounce off one another rapidly and freely
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3 years ago
A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What
Gemiola [76]

Answer:

The answer to your question:  0.7 M

Explanation:

Data

V of KOH = 90 ml

[KOH] = ?

V H2SO4 = 21.2 ml

[H2SO4] = 1.5 M

                       2KOH(aq)  +  H₂SO₄(aq)   →   K₂SO₄(aq)  +  2H₂O(l)

Molarity = moles / volume

moles of H₂SO₄ = (1.5) (21.2)

                           = 31.8

                    2 moles of KOH --------------  1 mol of H₂SO₄

                   x                           --------------  31.8 mol of H₂SO₄

                    x = (31.8)(2) / 1

                    x = 63.8 moles of KOH

Molarity = 63.8 / 90

             = 0.7 M

7 0
3 years ago
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