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Shtirlitz [24]
3 years ago
9

What would be the freezing point of a 1.7 mole aqueous ethylene glycol solution? The freezing point depression constant for wate

r = 1.86 degrees Celsius per mole.
a) 3.2°C

b) – 1.1°C

c) 0.0°C

d) – 3.2°C
Chemistry
1 answer:
vaieri [72.5K]3 years ago
7 0

<u>Answer:</u>

The freezing point of a 1.7 mole aqueous ethylene glycol solution is 3.2°C

<u>Explanation:</u>

Moles present in aqueous ethylene glycol solution= m=1.7

Depression constant for water = K_b= 1.8 deg C/m

According to formula  

T_f=K_b*m

T_f=1.8* 1.7

=3.2

Hence the freezing point of aqueous ethylene glycol solution is 3.2°C

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Answer: 0.0508mL

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3 years ago
Because of friction, some work is always changed into
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Thermal Energy

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The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur
vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

T_1 = 100^0 C = 100+273 = 373 \ K \\ \\  T_2 = 113^0 C = 113 + 273 = 386 \ K

R_1 = \dfrac{1}{7}

R_2 = \dfrac{1}{49}

Thus; \dfrac{R_2}{R_1} = 7

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})

1.9459 = \dfrac{Ea}{8.314}* 9.0292  *10^{-5}

1.9459*8.314 = Ea * 9.0292*10^{-5}

16.1782126= Ea * 9.0292*10^{-5}

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b)

here;

T_2 = 386 \  K  \\ \\T_1 = (89.8 + 273)K = 362.8 \ K

In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})

In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})

In (\dfrac{R_2}{R_1}) = 0.00357

\dfrac{R_2}{R_1}= e^{0.00357}

\dfrac{R_2}{R_1}= 1.0035

where ;

R_2 = \dfrac{1}7{}

R_1 = \dfrac{1}{t}

Now;

\dfrac{t}{7}= 1.0035

t = 7.0245 mins

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