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worty [1.4K]
2 years ago
10

A sample of oxygen gas occupies a volume of 160 liters at 364 K. What will be the volume of the gas when the temperature drops t

o 273 K?
210 L

120 L

62 L

470 L
Chemistry
1 answer:
Ivahew [28]2 years ago
8 0

Answer:

62L

.

.

.

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tysm.

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hope it help

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In the third period of the periodic table sodium is followed by magnesium aluminum silicon and phosphorus which of these element
wariber [46]
Answer:
            Phosphorous has the smallest atomic size.

Explanation:
                   As we know these elements belong to same period means there valence shell is the same. So moving from left to right along the period the shell number remains constant but the number of protons and electrons increases. So, due to increase in number of protons the nuclear charge increases hence attracts the valence electrons more effectively resulting in the decrease of atomic size.

Elements and their atomic radius are as follow,

<span><span>Magnesium          0.160 nm
</span><span>
Aluminium           0.130 nm
</span><span>
Silicon                  0.118 nm
</span><span>
Phosphorus         <span>0.110 nm</span></span></span>
6 0
3 years ago
What process is being shown by water being given off from each bond site?
34kurt
The process that is being shown by water being given off from a bond site is DEHYDRATION SYNTHESIS.
Dehydration synthesis is the process of joining two molecules or compounds together as a result of removal of water.
4 0
3 years ago
The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure
Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature T_1

P_2 = vapor pressure at temperature T_2

\Delta H_{vap} = Enthalpy of vaporization  

R = Gas constant = 8.314 J/mol K

1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

The vapor pressure at temperature 363 K is 0.6970 atm

2) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =373 K

T_2 = final temperature =383 K

P_1=1 atm, P_2?

Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

The vapor pressure at 383 K is 1.410 atm

8 0
3 years ago
How is density found
Yanka [14]
Through mass and volume. 
3 0
2 years ago
Read 2 more answers
The particles of a gas inside a balloon are experiencing an increase in the average kinetic energy and the number of collisions
Sever21 [200]

Answer:

1. Higher gas pressure inside the balloon.

Explanation:

An increase in the average kinetic energy and the number of collisions represents increases on gas temperature and pressure inside the ballon. Hence, the answer is 1.

6 0
3 years ago
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