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olasank [31]
3 years ago
9

Please give answers from h

Chemistry
1 answer:
Lina20 [59]3 years ago
8 0

Answer:

option b

Explanation:

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A person weighs a different amount on other surfaces in space than they do on Earth because the gravitational pull of each surfa
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A. The person weighs 56 pounds more on Mars than on the moon

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Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of glucose.
Rainbow [258]

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

4 0
2 years ago
Chemistry Stoichiometry Question
Zielflug [23.3K]

The correct answer is 1.1 moles

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Answer:D

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