At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Let's break down each question with a detailed, step-by-step solution:
### 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
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
