Answer:
Explanation:
1) Convert 11.03 moles of calcium nitrate to grams.
Given data:
Number of moles = 11.03 mol
Mass of calcium nitrate = ?
Solution:
Formula:
Number of moles = mass / molar mass
Now we will rearrange the formula because we have to calculate the mass.
Mass = number of moles × molar mass
Molar mass of calcium nitrate = 164.088 g/mol
Mass = 11.03 mol × 164.088 g/mol
Mass = 1809.89 g
2. How many molecules are contained in 103.4g of sulfuric acid?
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Given data:
Number of molecules = ?
Mass of sulfuric acid = 103.4 g
Solution:
Formula:
Number of moles = mass / molar mass
Number of moles = 103.4 g/ 98.079 g/mol
Number of moles = 1.05 mol
1 mole = 6.022 × 10²³ molecules
1.05 mol × 6.022 × 10²³ molecules / 1mol
6.323 × 10²³ molecules of sulfuric acid
3. 3.25 x 1024 molecules of dinitrogen pentoxide would be how many moles?
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Given data:
Number of molecules of dinitrogen pentoxide = 3.25 × 10²⁴
Moles of dinitrogen pentoxide = ?
Solution:
1 mole = 6.022 × 10²³ molecules
1 mol × 3.25 × 10²⁴ molecules / 6.022 × 10²³ molecules
5.4 moles
4. What would be the mass of 9.03 x 1021 molecules of hydrobromic acid?
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Given data:
Number of molecules of hydrobromic acid = 9.03 × 10²¹
Mass of hydrobromic acid = ?
Solution:
First of all we will calculate the moles of hydrobromic acid.
1 mole = 6.022 × 10²³ molecules
1 mol × 9.03 × 10²¹ molecules / 6.022 × 10²³ molecules
0.015 moles
Mass of hydrobromic acid:
Mass = number of moles × molar mass
Mass = 0.015 mol × 80.9 g/mol
Mass =1.21 g
5. A sample of iron (III) chloride has a mass of 26.29g. How many moles would this be?
Given data:
Number of moles =?
Mass of iron(III) chloride = 26.29 g
Solution:
Formula:
Number of moles = mass / molar mass
Molar mass of iron chloride = 162.2 g/mol
Now we will put the values in formula.
Moles = 26.29 g/ 162.2 g/mol
Moles = 0.16 g
1. Converting 11.03 moles of calcium nitrate to grams gives 1,809.89 grams.
2. The number of molecules that are contained in 103.4 g of sulfuric acid is [tex]6.35 \times 10^{23}[/tex] molecules.
3. Converting [tex]3.25 \times 10^{24}[/tex] molecules of dinitrogen pentoxide to moles is equal to 5.3987 moles.
4. The mass of [tex]9.03 \times 10^{21}[/tex] molecules of hydrobromic acid is 1.2135 grams.
5. The number of moles contained in 26.29 g of iron (III) chloride is 0.1621 moles.
Given the following data:
Scientific data:
1. To convert 11.03 moles of calcium nitrate to grams:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 11.03 \times 164.088[/tex]
Mass of calcium nitrate = 1,809.89 grams.
2. To calculate how many molecules are contained in 103.4 g of sulfuric acid:
First of all, we would determine the number of moles contained in 103.4 g of sulfuric acid.
[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{103.4}{98.079}[/tex]
Number of moles = 1.0543 moles
By stoichiometry:
1 mole of sulfuric acid = [tex]6.02 \times 10^{23}[/tex] molecules
1.0543 mole of sulfuric acid = X molecules
Cross-multiplying, we have:
[tex]X = 1.0543 \times 6.02 \times 10^{23}[/tex]
X = [tex]6.35 \times 10^{23}[/tex] molecules
3. To convert [tex]3.25 \times 10^{24}[/tex] molecules of dinitrogen pentoxide to moles:
By stoichiometry:
1 mole of dinitrogen pentoxide = [tex]6.02 \times 10^{23}[/tex] molecules
X mole of dinitrogen pentoxide = [tex]3.25 \times 10^{24}[/tex] molecules
Cross-multiplying, we have:
[tex]X=\frac{3.25 \times 10^{24}}{6.02 \times 10^{23}}[/tex]
X = 5.3987 moles
4. To calculate the mass of [tex]9.03 \times 10^{21}[/tex] molecules of hydrobromic acid:
First of all, we would determine the number of moles contained in [tex]9.03 \times 10^{21}[/tex] molecules of hydrobromic acid.
By stoichiometry:
1 mole of hydrobromic acid = [tex]6.02 \times 10^{23}[/tex] molecules
X mole of hydrobromic acid = [tex]9.03 \times 10^{21}[/tex] molecules
Cross-multiplying, we have:
[tex]X=\frac{9.03 \times 10^{21} }{6.02 \times 10^{23}}[/tex]
X = 0.015 moles
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 0.015 \times 80.91[/tex]
Mass of hydrobromic acid = 1.2135 grams.
5. To calculate the number of moles contained in 26.29 g of iron (III) chloride:
[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{26.29}{162.2}[/tex]
Number of moles = 0.1621 moles.
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