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lora16 [44]
2 years ago
13

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

Chemistry
1 answer:
Rainbow [258]2 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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The new pressure of the gas that initially have a pressure of 853.0 millibars at a temperature of 29.0 °C is 1011.17 millibars. Details about pressure can be found below.

<h3>How to calculate pressure?</h3>

The pressure of a given gas can be calculated using the following formula:

P1/T1 = P2/T2

Where;

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853/302 = P2/358

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305374 = 302P2

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Therefore, the new pressure of the gas that initially have a pressure of 853.0 millibars at a temperature of 29.0 °C is 1011.17 millibars.

Learn more about pressure at: brainly.com/question/15175692

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The more electronegative an atom is the more unwilling it is to lose its electrons in a compound. If you do try to take a very EN atom away from a compound you'll need to apply a lot of energy for that to happen. I can give an example of a single atom though 

<span>Cl has 7 valence electron filled and every atom wants to be like nobles (noble gases), so it's not going to give an electron away b/c it's really close to being like a noble gas. Noble gases are the most stable atoms, which is why I say stability counts.</span>
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