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### 1. Analyzing Samples of Magnesium Sulfide (MgS)
A sample of magnesium sulfide (MgS) is analyzed.
- First Sample:
- 12 g of magnesium (Mg)
- 16 g of sulfur (S)
The ratio of magnesium to sulfur in the first sample can be calculated as follows:
[tex]\[ \frac{\text{Mass of S}}{\text{Mass of Mg}} = \frac{16}{12} = \frac{4}{3} \][/tex]
- Second Sample:
- 36 g of magnesium (Mg)
- Let [tex]\( x \)[/tex] be the mass of sulfur (S)
Using the same ratio as the first sample:
[tex]\[ \frac{x}{36} = \frac{4}{3} \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 36 \times \frac{4}{3} = 36 \times 1.3333 = 48 \][/tex]
So, the value of [tex]\( x \)[/tex] is 48 g of sulfur.
### 2. Extracting Silver from Silver Bromide
- The atomic mass of silver (Ag) is 108.
- The atomic mass of bromine (Br) is 80.
- The combined molecular mass of silver bromide (AgBr) is:
[tex]\[ 108 + 80 = 188 \][/tex]
We are given 564 g of silver bromide (AgBr), and we need to find out how much pure silver (Ag) can be extracted from it.
The fraction of the mass of Ag in AgBr is:
[tex]\[ \frac{108}{188} \][/tex]
The total mass of silver that can be extracted from 564 g of silver bromide is:
[tex]\[ 564 \times \frac{108}{188} = 324 \][/tex]
So, the mass of silver that can be extracted is 324 g.
### 3. Fixed Ratios of Atoms in Given Compounds
Let's evaluate the fixed ratio of atoms in each given compound:
3.1 CaCl₄:
- 1 calcium (Ca) atom
- 4 chlorine (Cl) atoms
The ratio [tex]\( \text{Ca}:\text{Cl}_4 \)[/tex] is:
[tex]\[ 1 : 4 \][/tex]
3.2 NaRg:
- 1 sodium (Na) atom
- 1 radon (Rg) atom
The ratio [tex]\( \text{Na}:\text{Rg} \)[/tex] is:
[tex]\[ 1 : 1 \][/tex]
3.3 NaO₄:
- 1 sodium (Na) atom
- 4 oxygen (O) atoms
The ratio [tex]\( \text{Na}:\text{O}_4 \)[/tex] is:
[tex]\[ 1 : 4 \][/tex]
3.4 CH₄:
- 1 carbon (C) atom
- 4 hydrogen (H) atoms
The ratio [tex]\( \text{C}:\text{H}_4 \)[/tex] is:
[tex]\[ 1 : 4 \][/tex]
3.5 AH₃S:
- 1 atom of element A
- 3 hydrogen (H) atoms
- 1 sulfur (S) atom
The ratio [tex]\( A:\text{H}_3 \times \text{S} \)[/tex] is:
[tex]\[ 1 : 3 \times 1 \][/tex]
or more simply,
[tex]\[ 1 : 3 \][/tex]
3.6 NH₃:
- 1 nitrogen (N) atom
- 3 hydrogen (H) atoms
The ratio [tex]\( \text{N}:\text{H}_3 \)[/tex] is:
[tex]\[ 1 : 3 \][/tex]
### Summary of Results
1. The value of [tex]\( x \)[/tex] is 48 g.
2. The mass of silver extracted from 564 g of silver bromide is 324 g.
3. Fixed ratios of atoms in given compounds:
- 3.1 CaCl₄: 1 : 4
- 3.2 NaRg: 1 : 1
- 3.3 NaO₄: 1 : 4
- 3.4 CH₄: 1 : 4
- 3.5 AH₃S: 1 : 3
- 3.6 NH₃: 1 : 3
### 1. Analyzing Samples of Magnesium Sulfide (MgS)
A sample of magnesium sulfide (MgS) is analyzed.
- First Sample:
- 12 g of magnesium (Mg)
- 16 g of sulfur (S)
The ratio of magnesium to sulfur in the first sample can be calculated as follows:
[tex]\[ \frac{\text{Mass of S}}{\text{Mass of Mg}} = \frac{16}{12} = \frac{4}{3} \][/tex]
- Second Sample:
- 36 g of magnesium (Mg)
- Let [tex]\( x \)[/tex] be the mass of sulfur (S)
Using the same ratio as the first sample:
[tex]\[ \frac{x}{36} = \frac{4}{3} \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 36 \times \frac{4}{3} = 36 \times 1.3333 = 48 \][/tex]
So, the value of [tex]\( x \)[/tex] is 48 g of sulfur.
### 2. Extracting Silver from Silver Bromide
- The atomic mass of silver (Ag) is 108.
- The atomic mass of bromine (Br) is 80.
- The combined molecular mass of silver bromide (AgBr) is:
[tex]\[ 108 + 80 = 188 \][/tex]
We are given 564 g of silver bromide (AgBr), and we need to find out how much pure silver (Ag) can be extracted from it.
The fraction of the mass of Ag in AgBr is:
[tex]\[ \frac{108}{188} \][/tex]
The total mass of silver that can be extracted from 564 g of silver bromide is:
[tex]\[ 564 \times \frac{108}{188} = 324 \][/tex]
So, the mass of silver that can be extracted is 324 g.
### 3. Fixed Ratios of Atoms in Given Compounds
Let's evaluate the fixed ratio of atoms in each given compound:
3.1 CaCl₄:
- 1 calcium (Ca) atom
- 4 chlorine (Cl) atoms
The ratio [tex]\( \text{Ca}:\text{Cl}_4 \)[/tex] is:
[tex]\[ 1 : 4 \][/tex]
3.2 NaRg:
- 1 sodium (Na) atom
- 1 radon (Rg) atom
The ratio [tex]\( \text{Na}:\text{Rg} \)[/tex] is:
[tex]\[ 1 : 1 \][/tex]
3.3 NaO₄:
- 1 sodium (Na) atom
- 4 oxygen (O) atoms
The ratio [tex]\( \text{Na}:\text{O}_4 \)[/tex] is:
[tex]\[ 1 : 4 \][/tex]
3.4 CH₄:
- 1 carbon (C) atom
- 4 hydrogen (H) atoms
The ratio [tex]\( \text{C}:\text{H}_4 \)[/tex] is:
[tex]\[ 1 : 4 \][/tex]
3.5 AH₃S:
- 1 atom of element A
- 3 hydrogen (H) atoms
- 1 sulfur (S) atom
The ratio [tex]\( A:\text{H}_3 \times \text{S} \)[/tex] is:
[tex]\[ 1 : 3 \times 1 \][/tex]
or more simply,
[tex]\[ 1 : 3 \][/tex]
3.6 NH₃:
- 1 nitrogen (N) atom
- 3 hydrogen (H) atoms
The ratio [tex]\( \text{N}:\text{H}_3 \)[/tex] is:
[tex]\[ 1 : 3 \][/tex]
### Summary of Results
1. The value of [tex]\( x \)[/tex] is 48 g.
2. The mass of silver extracted from 564 g of silver bromide is 324 g.
3. Fixed ratios of atoms in given compounds:
- 3.1 CaCl₄: 1 : 4
- 3.2 NaRg: 1 : 1
- 3.3 NaO₄: 1 : 4
- 3.4 CH₄: 1 : 4
- 3.5 AH₃S: 1 : 3
- 3.6 NH₃: 1 : 3
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