Given,
P1 = 0.98 atm
V1 = 0.5 L
V2 = 1.0 L
P2 = ?
Solution,
According to Boyle's Law,
P1V1 = P2V2
0.98 × 0.5 = 1.0 × P2
P2 = 0.98 × 0.5 × 1.0
P2 = 0.49 atm
Answer - The new pressure is 0.49 atm.
Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K
Explanation :
Formula used :

where,
= change in entropy
= change in enthalpy of fusion = 3.17 kJ/mol
As we know that:

= freezing point temperature = 
Now put all the given values in the above formula, we get:



Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K
Considering the Charles' law, the gas would have a temperature of -109.2 C.
<h3>Charles' law</h3>
Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.
Charles' law is expressed mathematically as:

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

<h3>Temperature of the gas in this case</h3>
In this case, you know:
- P1= 1800 psi
- V1= 10 L
- T1= 20 C= 293 K (being 0 C= 273 K)
- P2= 1800 psi
- V2= 6 L
- T2= ?
You can see that the pressure remains constant, so you can apply Charles's law.
Replacing in the Charles's law:

Solving:


<u><em>T2=163.8 K= -109.2 C</em></u>
The gas would have a temperature of -109.2 C.
Learn more about Charles's law:
brainly.com/question/4147359?referrer=searchResults
Group 1 elements (usually called alkali metals) are not very electronegative and have small ionization energies due to that. The reason why they are not very electronegative is that they really want to loose their one valence electron so that they can have a noble gas electron configuration (completed octet).
I hope this helps.
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x <u> 1 mole CH4 </u> = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.