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3 years ago

How does the number of molecules in 1 mol of oxygen compare with the number of molecules in 1 mol of nitrogen?

Chemistry

1 answer:

frozen [14]3 years ago

7 0

Hey there!

The number of molecules is the same.

One mole of a type of molecule is <u><em>always</em></u> 6.022 x 10²³ molecules, so no matter what the type of molecule is there will be the same amount of molecules.

The mass, volume, or other properties can be different, but the number of molecules will be the same.

Hope this helps!

You might be interested in

How many moles of MgO are produced when .250 mol of Mg reacts completely with O2

nignag [31]

Answer:

0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂

Explanation:

In first place, the balanced reaction between Mg and O₂ is:

2 Mg + O₂ ⇒ 2 MgO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

Mg: 2 moles
O₂: 1 mole
MgO: 2 moles

Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?

$moles of MgO=\frac{0.250 moles of Mg*2 moles of MgO}{2 moles of Mg}$

moles of MgO= 0.250

<u><em>0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂</em></u>

3 0

3 years ago

How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2

lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0

2 years ago

Use the Bohr Model below to answer the questions:

Bas_tet [7]

Answer:

Ca(Calcium)

20 electrons

2 valence electrons

Explanation:

3 0

2 years ago

How is inoic and molecular solids differ from each other in solution

Irina18 [472]

Molecular are only between non-metals. Ionic has a higher melting point

4 0

3 years ago

Read 2 more answers

Muriatic acid, HCl, is often used to remove rust. A solution of muriatic acid, HCl, reacts with Fe2O3 deposits on industrial equ

Alina [70]

Answer:

Explanation:

Step 1:

The balanced equation for the reaction.

Fe2O3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2O

Step 2 :

Determination of the masses of HCl and Fe2O3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Fe2O3 = 159.70g/mol

Molar mass of HCl = 36.46 g/mol

Mass of HCl from the balanced equation = 6 x 36.46 = 218.76g

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl

Step 3:

Determination of the mass of HCl needed to react with 439g of Fe2O3. This is illustrated below:

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl.

Therefore, 439g of Fe2O3 will react with = (439 x 218.76) /159.70 = 601.35g of HCl.

Step 4:

Conversion of 601.35g of HCl to mole. This is illustrated below:

Molar mass of HCl = 36.46 g/mol

Mass of HCl = 601.35g

Number of mole = Mass/Molar Mass

Number of mole of HCl = 601.35/36.46

Number of mole of HCl = 16.49 moles

Step 5:

Determination of the volume of the HCl that reacted.

This is illustrated below:

Mole of HCl = 16.49 moles

Molarity of HCl = 5.50 M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 16.49/5.5

Volume of HCl = 3L

Therefore the volume of HCl needed for the reaction is 3L

5 0

2 years ago