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aleksandrvk [35]
3 years ago
14

A glycerol with three fatty acids attached is referred to as a ______.

Chemistry
1 answer:
icang [17]3 years ago
5 0

 

 

 

<span>Triglyceride is an ester that is derived from a glycerol with three fatty acids attached to it.  It is also known as triacylglyceride and triacylglycerol. Triglyceride are the main components of body fat in humans, some other animals and vegetable as well. </span>


Answer: Triglyceride or Fat

 

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3 years ago
Which of the following samples would be the most compressible?
Andrej [43]

Answer:

CO2 (g)

Explanation:

In solids the interatomic or intermolecular space is least . It is most pronounced in gases . That is why inter molecular or interatomic attraction is least in gases . That is why gas flows .

Hence , when we try to compress a gas , due to inter molecular space , it is most likely to get compressed . It will be least compressed when we try to compress a solid because of lack of intermolecular space .

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3 years ago
A 48.3 mL sample of gas in a cylinder is warmed from 22 C, 87 C to 0.0 C what is the final temperature
snow_lady [41]

The answer is:

the volume stays the same. it is the pressure that increases

7 0
3 years ago
One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

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3 years ago
Q1.
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