Answer : The partial pressure of the 
in the tank in psia is, 32.6 psia.
Explanation :
As we are given 75 % and 25 % 
in terms of volume.
First we have to calculate the moles of and .
Now we have to calculate the mole fraction of .
Now we have to calculate the partial pressure of the gas.
conversion used : (1 Kpa = 0.145 psia)
Therefore, the partial pressure of the in the tank in psia is, 32.6 psia.
Answer:
See explanation
Explanation:
Extraction has to do with the separation of the components of a mixture by dissolving the mixture in a set up involving two phases. One phase is the aqueous phase (beneath) while the other is the organic phase (on top). The solvents used for the two phases must not be miscible. Water commonly is used for the aqueous phase.
Ethanol is an important solvent in chemistry but the solvent is miscible with water in all proportions. As a result of this, ethanol is a poor solvent for carrying out extraction.
Solutions are said to be C. homogeneous mixtures, composed of two or more substances. It is usually liquid, however it may be solid or gas.
Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.
To increase the energy of the emitted electrons, the frequency of the incident light on the metal must be increased.
<h3>What is energy of emitted electron?</h3>
The maximum energy of an emitted electron is equal to the energy of a photon for frequency f (E = hf ), minus the energy required to eject an electron from the metal's surface, also known as work function.
Ee = E - W
<h3>Energy of the emitted electron</h3>
The energy of emitted electrons based on the research of Albert Einstein is given as;
E = hf
where;
- h is planck's constant
- f is frequency of incident light on the metal
Thus, to increase the energy of the emitted electrons, the frequency of the incident light on the metal must be increased.
Learn more about energy of electron here: brainly.com/question/11316046
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