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lora16 [44]
3 years ago
13

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of glucose.

Chemistry
1 answer:
Rainbow [258]3 years ago
4 0

Answer:

-0.1767°C (Option A)

Explanation:

Let's apply the colligative property of freezing point depression.

ΔT = Kf . m. i

i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1

m = molality (mol of solute / 1kg of solvent)

We have this data → 0.095 m

Kf is the freezing-point-depression constantm 1.86 °C/m, for water

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

(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1

T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C

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Answer:

The answer is

<h2>720 Joules</h2>

Explanation:

The kinetic energy of a body can be found by using the formula

<h3>K.E =  \frac{1}{2}  {mv}^{2}</h3>

where

m is the mass

v is the velocity / speed

From the question

mass = 10 kg

velocity = 12 m/s

Substitute the values into the above formula and solve

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<h3>KE =  \frac{1}{2}  \times 10 \times  {12}^{2}  \\   = 5 \times 144  \:  \:</h3>

We have the final answer as

<h3>720 Joules</h3>

Hope this helps you

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3 years ago
Initially a beaker contains 225.0 ml of a 0.350 M MgSO4 solution. Then 175.0 ml of water are added to the beaker. Find the conce
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4 years ago
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If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the re
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