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1.3 Which one of the following expressions gives the percentage of nitrogen (N) in calcium nitrate, [tex]Ca(NO_3)_2[/tex]?

A. [tex]\frac{14}{164} \times 100[/tex]
B. [tex]\frac{28}{164} \times 100[/tex]
C. [tex]\frac{14}{102} \times 100[/tex]
D. [tex]\frac{28}{102} \times 100[/tex]

1.4 Consider the balanced chemical equation below:
[tex]\[
N_2 + 3H_2 \longrightarrow 2NH_3
\][/tex]

What volume of [tex]NH_3[/tex] will be produced when [tex]2 \, \text{dm}^3[/tex] of [tex]N_2[/tex] reacts completely?

A. [tex]2 \, \text{dm}^3[/tex]
B. [tex]4 \, \text{dm}^3[/tex]
C. [tex]22.4 \, \text{dm}^3[/tex]
D. [tex]44.8 \, \text{dm}^3[/tex]

1.5 Which one of the samples of gases below will have the largest number of molecules if the mass of each of the samples is the same?

A. [tex]H_2O[/tex]
B. [tex]N_2[/tex]
C. [tex]CO_2[/tex]
D. [tex]Cl_2[/tex]

Sagot :

GinnyAnsweridn
Let's break down each part of the problem and solve it step-by-step:

### 1.3 Percentage of Nitrogen in Calcium Nitrate ([tex]$Ca(NO_3)_2$[/tex])

First, we need to understand the molecular composition of calcium nitrate:
- Calcium (Ca): 1 atom with an atomic weight of 40
- Nitrogen (N): 2 atoms with an atomic weight of 14 each
- Oxygen (O): 6 atoms with an atomic weight of 16 each

The formula weight of calcium nitrate can be calculated as follows:
[tex]\[ \text{Total formula weight} = (1 \times 40) + (2 \times 14) + (6 \times 16) = 40 + 28 + 96 = 164 \][/tex]

The expression for the percentage of nitrogen in the compound is given by:
[tex]\[ \text{Percentage of nitrogen} = \frac{\text{Total weight of nitrogen in the compound}}{\text{Total formula weight}} \times 100 \][/tex]

Since there are 2 nitrogen atoms:
[tex]\[ \text{Total weight of nitrogen in the compound} = 2 \times 14 = 28 \][/tex]

Thus, the correct expression is:
[tex]\[ \frac{28}{164} \times 100 \][/tex]

Hence, the correct answer is:
[tex]\[ \boxed{\text{B. } \frac{28}{164} \times 100} \][/tex]

### 1.4 Volume of Ammonia ([tex]$NH_3$[/tex]) Produced

Given the balanced chemical equation:
[tex]\[ N_2 + 3H_2 \rightarrow 2NH_3 \][/tex]

This tells us that 1 volume of nitrogen gas ([tex]$N_2$[/tex]) produces 2 volumes of ammonia ([tex]$NH_3$[/tex]).

If we start with 2 dm[tex]\(^3\)[/tex] of nitrogen gas ([tex]$N_2$[/tex]), it will produce:
[tex]\[ 2 \text{ dm}^3 \times 2 = 4 \text{ dm}^3 \][/tex]

Therefore, the volume of ammonia produced is:
[tex]\[ \boxed{\text{B. } 4 \text{ dm}^3} \][/tex]

### 1.5 Gas with the Largest Number of Molecules

To determine which gas has the largest number of molecules when their masses are the same, we must compare their molar masses:
- [tex]\(H_2O\)[/tex]: 18 g/mol
- [tex]\(N_2\)[/tex]: 28 g/mol
- [tex]\(CO_2\)[/tex]: 44 g/mol
- [tex]\(Cl_2\)[/tex]: 70.9 g/mol

The gas with the smallest molar mass will have the largest number of molecules because, for a given mass, the number of moles is inversely proportional to the molar mass.

Among the listed gases:
[tex]\[ H_2O \text{ has the smallest molar mass of 18 g/mol} \][/tex]

Therefore, the gas with the largest number of molecules is:
[tex]\[ \boxed{\text{A. } H_2O} \][/tex]

By following these steps, we've arrived at the correct answers for each part of the problem!
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