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lana66690 [7]
3 years ago
12

a car is traveling at a speed of 55 mph the speed increased to 75 mph if the time needed to hange its speed is 4 seconds whats t

he acceleration ?
Chemistry
1 answer:
Savatey [412]3 years ago
6 0

v = 75 mph = 33.52 m/s

u = 55 mph = 24.58 m/s

t = 4 sec

acceleration = v-u/t

= 33.52-24.58/4

= 8.94/4

= 2.235 m/s

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

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To calculate the molality of solution, we use the equation:

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Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

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\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

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Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

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Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

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\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

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m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

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