contestada

How many atoms are in a sample of copper with a mass of 18.46 grams?
2.How many formula units are in a sample of salt with a mass of 67.69 grams?
3.How many molecules are in a sample of water with a mass of 44.99 grams?
4.What is the mass in grams of 0.250 moles Mg? 5.How many moles are there in 48.096 grams of sulfur?
6.What is the mass of 24.6 formula units of magnesium oxide?
7.How many molecules of dinitrogen pentoxide are contained in 123.46 grams of dinitrogen pentoxide?

Respuesta :

MissPhiladelphia
Just solve this by using the molar mass/molecular weight, and the avogadros number in a dimensional analysis
18.46 g Cu x (1 mole/64 g) x (6.022 x 10^23 atoms/1 mole) = 1.74 x 10^23 atoms
67.69 g NaCl x (1 mole/58 g) x (6.022 x 10^23 formula units/ 1 mole) = 7.03 x 10^23 formula units
44.99 g H2O x (1 mole/18 g) x (6.022 x 10^23 molecules/ 1 mole) = 1.51 x 10^24 molecules
0.250 moles Mg x (24 g/1 mole) = 6 g
48.096 g S x (1 mole/32 g) = 1.5 g
Dejanras

1) The answer is: 1.75·10²³ are in a sample of copper.

m(Cu) = 18.46 g; mass of copper.

n(Cu) = m(Cu) ÷ M(Cu).

n(Cu) = 18.46 g ÷ 63.55 g/mol.

n(Cu) = 0.29 mol; amount of copper.

Na = 6.022·10²³ 1/mol; Avogadro number.

N(Cu) = n(Cu) · Na.

N(Cu) = 0.29 mol · 6.022·10²³ 1/mol.

N(Cu) = 1.75·10²³; number of copper atoms.

2) The answer is: number of formula units is 6.975·10²³.

m(NaCl) = 67.69 g; mass of salt.

n(NaCl) = m(NaCl) ÷ M(NaCl).

n(NaCl) = 67.69 g ÷ 58.44 g/mol.

n(NaCl) = 1.16 mol; amount of salt.

Na = 6.022·10²³ 1/mol; Avogadro number.

N(NaCl) = n(NaCl) · Na.

N(NaCl) = 1.16 mol · 6.022·10²³ 1/mol.

N(NaCl) = 6.975·10²³; number of formula units.

3) The answer is: 1.506·10²⁴ molecules of water.

m(H₂O) = 44 g; mass of water.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 44.99 g ÷ 18 g/mol.

n(H₂O) = 2.5 mol; amount of water.

Na = 6.022·10²³ 1/mol; Avogadro number.

N(H₂O) = n(H₂O) · Na.

N(H₂O) = 2.5 mol · 6.022·10²³ 1/mol.

N(H₂O) = 1.506·10²⁴; number of water molecules.

4) The answer is: the mass of magnesium is 6.08 grams.

n(Mg) = 0.250 mol; amount of magnesium.

M(Mg) = 24.305 g/mol; molar mass of magnesium.

m(Mg) = n(Mg) · M(Mg).

m(Mg) = 0.250 mol · 24.305 g/mol.

m(Mg) = 6.076 g; mass of magnesium.

Magnesium (Mg) is metal from 2. group of Periodic table of elements with atomic number 12.

5) The answer is: there are 1.5 moles of sulfur.

m(S) = 48.096 g; mass of sulfur.

M(S) = 32.065 g/mol; molar mass of sulfur.

n(S) = m(S) ÷ M(S).

n(S) = 48.096 g ÷ 32.065 g/mol.

n(S) = 1.50 mol; amount of sulfur.

Sulfur is a chemical element with symbol S and atomic number 16.

6) The answer is: the mass is 1.65·10⁻²¹ grams.

Ar(MgO) = 40.3; relative atomic mass of magnesium oxide.

m(MgO) = 24.6 · 40.3 · 1.66·10⁻²⁷ kg.

m(MgO) = 1645.7·10⁻²⁷ kg.

m(MgO) = 1.65·10⁻²¹ g; mass og magnesium oxide.

Ar is relative atomic mass (the ratio of the average mass of atoms of a chemical element to one unified atomic mass unit) of an element.

The unified atomic mass unit (amu) is a standard unit of atom mass.  

One unified atomic mass unit is approximately the mass of one nucleon (1.66·10⁻²⁷ kg).

7) The answer is: 6.88·10²³ molecules of dinitrogen pentoxide.

m(N₂O₅) = 123.46 g; mass of dinitrogen pentoxide.

M(N₂O₅) = 108.01 g/mol; molar mass of dinitrogen pentoxide.

n(N₂O₅) = m(N₂O₅) ÷ M(N₂O₅).

n(N₂O₅) = 123.46 g ÷ 108.01 g/mol.

n(N₂O₅) = 1.14 mol; amount of dinitrogen pentoxide.

N(N₂O₅) = 1.14 mol · 6.022·10²³ 1/mol.

N(N₂O₅) = 6.88·10²³; number of molecules.

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