All categories

3 years ago

Does the boiling point of a liquid depend of the amount of liquid

1 answer:

5 0

A liquid boils at a temperature at which its vapor pressure is equal to the pressure of the gas above it. The lower the pressure of a gas above a liquid, the lower the temperature at which the liquid will boil. Boiling point represents a phase change of matter. So no, the boiling poibt of a liquid does not depend on the amount of liquid

You might be interested in

What is fiber and fabric

Sidana [21]

Answer:

Fibre , it is a long thin strand or something like thread of material.

Fabric, it is a cloth material made by weaving or knitting threads together.

6 0

1 year ago

Read 2 more answers

What is the redox half equation for 3Ag2S + 2Al --> 6Ag + Al2, and identity which material is oxidized and which is reduced?

marta [7]

Answer:

Al is oxidized while Ag is reduced.

Explanation:

The complete molecular equation is;

3Ag2S + 2Al --> 6Ag + Al2S3

Oxidation half equation;

2Al ------> 2Al^3+ + 6e

Reduction half equation;

6Ag^+ + 6e -------> 6Ag

Overall redox reaction equation;

2Al + 6Ag^+ ----->2Al^3+ + 6Ag

Hence; Al is oxidized while Ag is reduced.

5 0

2 years ago

Why are alloys generally used to make everyday objects

dezoksy [38]

Chem bonds by exchanged charges create new characteristics as a result

4 0

2 years ago

What is the molecular geometry of PBr3? A. trigonal pyramidal B. trigonal planar C. linear D....

lubasha [3.4K]

The correct answer is:

TRIGONAL PYRAMIDAL

4 0

2 years ago

16. The concentration of a solution of potassium hydroxide is determined by titration with nitric

____ [38]

Answer:

$M_{base}=0.709M$

Explanation:

Hello,

In this case, since the reaction between potassium hydroxide and nitric acid is:

$KOH+HNO_3\rightarrow KNO_3+H_2O$

We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:

$n_{acid}=n_{base}$

That in terms of molarities and volumes is:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, solving the molarity of the base (KOH), we obtain:

$M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M$

Regards.

3 0

2 years ago