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4 years ago

Name the basic assumptions that the kinetic theory makes about gases

1 answer:

3 0

Some assumptions is that gas is a composed of large, identical, constantly moving in different directions, molecules. another is that they undergo flawless elastic collisions

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Which soil layer contains the most humus? A) A horizon B) B horizon C) C horizon D) D horizon

Bad White [126]

None. It's the O horizon. Compromises of mostly decomposed plant litter.

5 0

3 years ago

Consider the reaction for the combustion of butane (C4H10), a component of liquefied petroleum (LP) gas. 2C4H10(g)+13O2(g)→8CO2(

polet [3.4K]

Answer:

The answer to your question is:

a) 3.87 mol of CO2

b) 14.63 mol of O2

Explanation:

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

CO2 = ?

Butane 0.968 moles

From the balance equation 2 moles of butane ---------- 8 moles of CO2

0.968 moles of butane ------ x

x = (0.968 x 8) / 2

x = 3.87 moles of CO2

moles O2 = ?

butane = 2.25 mol

From the balance equation

2 moles of butane ---------- 13 mol O2

2.25 mol ---------- x

x = (2.25 x 13) / 2

x = 14.63 mol of O2

5 0

3 years ago

Viết công thức hóa học tạo bởi Mg và O

jasenka [17]

Phương trình hóa học: 2Mg+O2->2Mgo

3 0

3 years ago

I WILL RATE BRAINLIEST I JUST NEED HELP!!!!

Ierofanga [76]

15:15 which cancels to 1:1 so for each mole of co2 u get a mole of carbon monoxide

brainliest and thanks plz:)))))))))))

8 0

4 years ago

Read 2 more answers

Which of the following substances would you predict to have the highest DHvap? Group of answer choices CH3CH2CH2CH3 CH3CH2OH HF

Katyanochek1 [597]

Answer:

a) $CH_3CH_2CH_2CH_3$

Explanation:

In this question we have the following answer choices:

a) $CH_3CH_2CH_2CH_3$

b) $CH_3CH_2OH$

c) $HF$

d) $CH_3Cl$

e) $HOCH_2CH_2OH$

We have to remember the relationship between intermolecular forces and vapor pressure. If we have stronger intermolecular forces we will have less vapor pressure because the molecules have more interactions between them, so, the molecules will prefer to stay in a liquid state rather than a gaseous state. Now, we have to check each molecule:

a) $CH_3CH_2CH_2CH_3$ (Van der waals interactions)

b) $CH_3CH_2OH$ (Hydrogen bonding)

c) $HF$ (Hydrogen bonding)

d) $CH_3Cl$ (Dipole-dipole interaction)

e) $HOCH_2CH_2OH$ (Hydrogen bonding)

For molecules b, c and e we have hydrogen bond to a heteroatom (O, N, S, or P). In this case oxygen, therefore we will have hydrogen bonding interactions (a very strong interaction). So, we can discard these ones.

In molecule e, we have "Cl" bond to a "C" therefore we will have the presence of a dipole (due to the electronegativity difference). If we have a dipole, we will have a dipole-dipole interaction (a strong interaction, less than hydrogen bonding but still is a strong interaction).

In molecule a, we have only Van der Waals interactions because in this molecule we have only carbon and hydrogen atoms bonded by single bonds. So, we will have a non-polar molecule. These interactions are the weakest interactions of all the molecules given. So, if we have weaker interactions the molecules can be converted to a gas state more easily and we have more vapor pressure. 

7 0

3 years ago