Using distance and can help you identify the position of something

1) Question 1. A sample of pure calcium fluoride with a mass of 15.0 g contains 7.70 g of calcium. How much calcium is contained in 40.0 g of calcium fluoride?

Answer:

20.5 g Ca

Explanation:

Calcium fluoride being a pure substance, has a fixed chemical compositon, which means that their elements (atoms), in any sample, are always in the same proportion.

Then, you can set the followoing proportion:

7.70 g Ca / 15.0 g CaF₂ = x / 40.0 g CaF₂

Now, solve for x:

x = 40.0 g CaF₂ × 7.70 g Ca / 15.0 g CaF₂ = 20.5 g Ca.

The answer must be expressed with 3 significan figures, since the three data are given with 3 significan figures.

2) Question 2. Silver has two isotopes. One isotope contains 60 neutrons and has a percent abundance of 51.839% the other contains 62 neutrons. Give the atomic symbols for the two isotopes.

Answer:

Isotope that contains 60 neutrons: ¹⁰⁷₄₇Ag

Isotope that contains 62 neutrons: ¹⁰⁹₄₇Ag

Explanation:

Isotopes are atoms of the same element, with different number of neutrons, so all the isotopes of an element have the same atomic number, Z (number of protons) and different mass number A (number of neutrons + protons).

The atomic symbols for isotopes uses the chemical symbol of the atom (Ag for silver); a subscript to the left of the chemical symbol that represents the atomic mass (47 for silver); and a superscript to the left of the chemical symbol that represents the mass number.

The mass number, A, for each silver isotope is calculated in this way:

Isotope that contains 60 neutrons: A = 60 + 47 = 107

Isotope that contains 62 neutrons: A = 62 + 47 = 109

Then, the respective atomic symboles are:

Isotope with 60 neutrons: ¹⁰⁷₄₇Ag

Isotope with 62 neutrons: ¹⁰⁹₄₇Ag

3) Question 3. What is the average atomic weight of silver? What is the mass (in grams) of a single chlorine molecule?

Answer:

107.96 amu

Explanation:

The atomic wieght of the elements is the weighted average weight of the different isotopes of the same element found in the nature, taking into account their relative abundance.

Average atomic weight = ∑( abundance × individual atomic weight)

Average atomic weight of silver = (abundace of isotope 1 × atomic weight isotope 1) + (abundance isotope 2 × atomic weight isotope 2)

Abundance isotope 1 = 51.839%

Atomic weight isotope 1 = mass number = 60 neutrons + 47 protons = 107 a.m.u
Abundance isotope 2 = 100% - 51.839% = 48.161%

Atomic weight isotope 2 = 62 neutrons + 47 protons = 109 a.m.u.

Average atomic weight of silve = 107 amu × 51.839% + 109 amu × 48.161% = 107.96 amu

4) Question without number: What is the mass (in grams) of a single chlorine molecule

Answer:

1.177 × 10⁻²² g

Explanation:

The chlorine molecule is diatomic. That means that each molecule contains two atoms. The chemical formula of chlorine molecule is Cl₂.

The mass of one mole of molecules of Cl₂ is 2 × 35.453 g/mol = 70.906 g/mol

Mass of a single molecule = mass of one mol / number of molecules in one mol

Mass of a single molecule = 70.906 g/mol / (6.022 × 10²³ molecules / mol)

Mass of a single molecule = 11.77 × 10⁻²³ g = 1.177 × 10⁻²² g

5) Question 4. How many moles of ethanol (CH₃CH₂OH) are in 141 mg?

Answer:

0.00301 mol

Explanation:

Molar mass of CH₃CH₂OH = 2 × 12.011 g/mol + 6 × 1.008 g/mol + 15.999 g/mol = 46.069 g/mol.

Formula: number of moles = mass in grams / molar mass

Calculation: number of moles = 141 mg × ( 1 g /1000 mg) / (46.069 g/mol = 0.00301 mol

4 0

3 years ago