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tino4ka555 [31]
3 years ago
10

How many particles are in on mole?

Chemistry
1 answer:
shusha [124]3 years ago
6 0

The answer is 1023 particles

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A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A
anygoal [31]

this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.

c1v1 = c2v2

where c1 is concentration and v1 is volume of the concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL

substituting these values in the formula

1.50 M x 50.0 mL = C x 250 mL

C = 0.300 M

concentration of the final solution is A) 0.300 M

4 0
3 years ago
Read 2 more answers
The Law of Conservation of Matter describes how mass is not lost during a chemical reaction.
weeeeeb [17]

False, it states that matter can be nether created nor destroyed

7 0
2 years ago
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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
3 years ago
CaO + H2O -&gt; Ca(OH)2
yawa3891 [41]

The % yield of Ca(OH)₂ : 62.98%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

CaO + H₂O ⇒ Ca(OH)₂

mass CaO= 4.2 g

mol CaO(MW=56,0774 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{4.2}{56,0774 g/mol}\\\\mol=0.075

mol Ca(OH)₂ based on mol CaO

mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075

mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical

\tt mass=mol\times MW\\\\mass=0.075\times 74,093 g/mol\\\\mass=5.557~g

% yield :

\tt =\dfrac{actual}{theoretical}\times 100\%\\\\=\dfrac{3.5}{5.557}\times 100\%\\\\=62.98\%

8 0
2 years ago
Which atom would form a cation?<br> F<br> Na<br> N<br> P
sineoko [7]

Answer:

I believe Na

Explanation:

Copper, because it is the only metal out of all of them.

Therefore copper is the only element that can loose electrons to have a positive charge, it is the most likely to become a cation.

5 0
3 years ago
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