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Olenka [21]
2 years ago
13

If 35.0 g of CH₃OH (MM = 32.0 g/mol) are dissolved in 500.0 mL of solution, what is the concentration of CH₃OH in the solution?

Chemistry
1 answer:
Stels [109]2 years ago
6 0

Answer:

mlllllllllllllllllll lodsssss

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Explain how the effects of nuclear charge and the increasing number of core electrons in an atom oppose each other.
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As the number of electrons grows, they occupy outer shelss (farther from the nucleus). And the outer electrons will feel not only the atraction of the protons from the nucleus, but the repulsion of the inner electrons.

Then, we see that the increase of nuclear charge is opposed by the increase of core electrons, and the outer (valence) electrons are not so tied to the nucleus as the core electrons are.

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The more the number of core shells the greater the shielding effect experience by electros in the outermost shells.

The shielding effect, explains why the valence eletrons are more easily removed from the atom than core electrons, and also explains some  trends of the periodic table: variationof the size of the atoms in a row, the greater the shielding efect, the less the atraction force felt by the outermos electron, the farther they are and the larger the atom. 


3 0
3 years ago
Explain the Law of Conservation of Mass in terms of the atoms and bonds in a chemical reaction.
nasty-shy [4]

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no matter is destroyed or created, it merely changes form. In terms of atoms and bonds, there will be the same amount of atoms at the beginning of an experiment as the amount of atoms at the end of experiment. All that will have happened, is that during the reaction, bonds will have been broken and formed making new compounds. However, the amount of atoms remains exactly the same because matter can not be created or destroyed

Hope this helps!

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Why are atoms considered the building blocks life?
dmitriy555 [2]
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7 0
2 years ago
For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox
ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

2) Limiting reagent: NO

3) Ammount of excess reagent: m_{N2}=4.274 g

Explanation:

<u>The reaction </u>

2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)

Moles of nitrogen monoxide

Molecular weight: M_{NO}=30 g/mol

n_{NO}=\frac{m_{NO}}{M_{NO}}

n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol

Moles of hydrogen

Molecular weight: M_{H2}=2 g/mol

n_{H2}=\frac{m_{H2}}{M_{H2}}

n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}

m_{N2}=10.682 g N_2

2) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}

m_{N2}=4.274 g

8 0
4 years ago
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