Answer: Kinetic Energy of the atoms also increases.
Explanation: We are given that the temperature of the gas increases.
Relation between kinetic energy and temperature follows:

where, K = Average Kinetic energy
R = Gas constant
T = Temperature
= Avogadro's number
As seen from the relation above, the Kinetic energy of the gas is directly proportional to the temperature, hence as the temperature increases, kinetic energy of the atom also increases.
Answer:
Explanation:
If an atom has 13 electrons then it belongs to p-block of periodic table.
s level can accommodate 2 electrons.
p level can accommodate 6 electrons.
13 means 1s2 2s2 2p6 3s2 3p1.
As you can see there totally 5 sub-shells.
Total number of shells are 3(1,2,3).
Answer : The correct option is, (C) 0.675 M
Explanation :
Using neutralization law,

where,
= concentration of
= 13.5 M
= concentration of diluted solution = ?
= volume of
= 25.0 ml = 0.0250 L
conversion used : (1 L = 1000 mL)
= volume of diluted solution = 0.500 L
Now put all the given values in the above law, we get the concentration of the diluted solution.


Therefore, the concentration of the diluted solution is 0.675 M
Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17%
concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride