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kenny6666 [7]
2 years ago
7

All of the following are considered greenhouse gases except

Chemistry
1 answer:
soldi70 [24.7K]2 years ago
8 0
A.) Except O2, all the gases are considered as greenhouse gases. Hope this helps!
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How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?
raketka [301]
<h2><u>Answer:</u></h2>

n = 0.0989 moles

<h2><u>Explanation:</u></h2>

n = PV / RT

P = 2.09atm

V = 1.13L

R = 0.08206

T = 291K

Plug the numbers in the equation.

n = (2.09atm)(1.13L) / (0.08206)(291K)

n = 0.0989 moles

3 0
3 years ago
Tell how you could distinguish between the pair of compounds by their infrared spectra: compound a ch3ch2ch2n(ch3)2 and compound
maksim [4K]

Answer:-

b. compound b will show an nh absorption while compound a will not.

Explanation:-

For compound B CH3CH2CH2NH2 there are two N-H bonds present. So it will show NH absorption.

But in compound A CH3CH2CH2N(CH3)2 there are no N-H bonds. So it will not show NH absorption.

6 0
3 years ago
What is the relationship between the number of particles and the overall pressure?
Thepotemich [5.8K]

Answer:

The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle's law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules

Explanation:

5 0
3 years ago
When 2.16g of H2 reacts with excess O2 by the following equation, 258 kJ of heat are released. What is the change of enthalpy as
alekssr [168]

Answer:

-241 kJ/mol

Explanation:

Let's consider the reaction of hydrogen with excess oxygen to form water.

2 H₂ + O₂ ⟶ 2 H₂O

When 2.16g of hydrogen reacts with excess oxygen, 258 kJ of heat are released, that is, Q = -258 kJ. Considering that the molar mass of hydrogen is 2.02 g/mol, the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas is:

ΔH° = -258 kJ/2.16 g × (2.02 g/1.00 mol) = -241 kJ/mol

8 0
2 years ago
A gas mixture contains 0.800 mol of N2, 0.200 mol of H2, and 0.150 mol of CH4. What is the mole fraction of H2 in the mixture?
Blababa [14]

Answer:

0.1739

Explanation:

0.800 mol of N2

0.200 mol of H2

0.150 mol of CH4

Total moles of the mixture = 0.8 + 0.2 + 0.150 = 1.150 mol

Mole fraction of H2 = Number of moles of H2 /  Total moles

Mole Fraction = 0.2 / 1.150 = 0.1739

4 0
2 years ago
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