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zavuch27 [327]
2 years ago
8

Calculate each of the following quantities: (c) Mass (g) of 5.85 mol of glycerol (C₃H₈O₃)

Chemistry
1 answer:
Sergio039 [100]2 years ago
5 0

<u>538.73 g</u> is the mass of 5.85 mol of glycerol (C₃H₈O₃)

<h3>What is glycerol?</h3>

Glycerol is a simple polyol chemical, usually known as glycerine in British English and glycerin in American English. It is a sweet-tasting, colourless, odourless, viscous liquid that is non-toxic. Glycerides, a class of lipids, contain the glycerol backbone.

It is frequently utilised in FDA-approved wound and burn therapies due to its antibacterial and antiviral characteristics. On the other hand, it can also serve as a medium for bacterial cultivation. It can be used as a reliable indicator of liver illness.

Additionally, it is employed temporarily in the food sector as a sweetener and in drug formulations as a humectant. Glycerol is hygroscopic in nature and miscible with water due to its three primary hydroxyl groups.

Learn more about glycerol

brainly.com/question/1842081

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A gas is allowed to expand from a volume of 400 ml to 2000 ml at a constant temperature. If the the initial pressure is 3 atm, c
Lostsunrise [7]

Answer:

P_{2}) = 0.6 \; atm

Explanation:

<u>Given the following data;</u>

Initial volume = 400 mL

Final volume = 2000 mL

Initial pressure = 3 atm

To find the final pressure P2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

PV = K

P_{1}V_{1} = P_{2}V_{2}

Substituting into the equation, we have;

3 * 400 = P_{2}* 2000

1200 = 2000P_{2}

P_{2}) = \frac {1200}{2000}

P_{2}) = 0.6 \; atm

8 0
3 years ago
2. HOW MUCH HEAT IS REQUIRED TO BE RELEASED WHEN
jarptica [38.1K]

Answer: -112200J

Explanation:

The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of water vapour = 30.0g

C = 187 J/ G°C

Φ = (Final temperature - Initial temperature)

= 100°C - 120°C = -20°C

Then apply the formula, Q = MCΦ

Q = 30.0g x 187 J/ G°C x -20°C

Q = -112200J (The negative sign does indicates that heat was released to the surroundings)

Thus, -112200 joules of heat is released when cooling the superheated vapour.

5 0
4 years ago
Balanced equation of potassium chloride and ammonium nitrate; it’s types, net and observation
ASHA 777 [7]

Potassium chloride reacts with ammonium nitrate to give ammonium chloride and potassium nitrate.

This is a type of double displacement reaction. The balanced chemical equation can be represented as,

KCl(aq)+NH_{4}NO_{3}(aq)-->NH_{4}Cl(aq)+KNO_{3}(aq)

Total ionic equation for this reaction will be,

K^{+}(aq)+Cl^{-}(aq)+NH_{4}^{+}(aq)+NO_{3}^{-}(aq)--> K^{+}(aq)+NO_{3}^{-}(aq)+NH_{4}^{+}(aq)+Cl^{-}(aq)

There is no apparent reaction as this reaction is not accompanied by the formation of a gas or a solid precipitate. We cannot observe any visual reaction as there is not net reaction taking place. All the ions remain as spectator ions.

5 0
4 years ago
How many moles of hydrogen are in 4.12 moles of water (H 2 O) ?
exis [7]

Answer:

tbh i don't even know..

8 0
3 years ago
A solution is prepared by mixing 525 mL of ethanol with 597 mL of water. The molarity of ethanol in the resulting solution is 8.
Alenkinab [10]

Answer:

\Delta V = 234.736\,mL

Explanation:

The quantity of moles of ethanol in the solution is:

n_{C_{2}H_{5}OH} = \left(\frac{597\,mL}{1000\,mL} \right)\cdot \left(8.35\,\frac{mol}{L} \right)

n_{C_{2}H_{5}OH} = 4.985\,mol

The mass and volume of ethanol in the solution are, respectively:

m_{C_{2}H_{5}OH} = (4.985\,mol)\cdot \left(46.07\,\frac{g}{mol} \right)

m_{C_{2}H_{5}OH} = 229.658\,g

V_{C_{2}H_{5}OH} = \frac{229.658\,g}{0.7893\,\frac{g}{mL} }

V_{C_{2}H_{5}OH} = 290.964\,mL

The difference between the total volume of water and ethanol mixed to prepare the solution and the actual volume of solution is:

\Delta V = (525\,mL+597\,mL) - (597\,mL + 290.964\,mL)

\Delta V = 234.736\,mL

8 0
3 years ago
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