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Sagot :
Certainly! Let's determine the symbol of the unknown metal [tex]\( M \)[/tex] through a step-by-step calculation:
### Step-by-Step Calculation:
1. Volume and Concentration of Sodium Hydroxide Solution:
- Volume of NaOH solution: [tex]\( 25 \, \text{cm}^3 \)[/tex]
- Convert volume from [tex]\( \text{cm}^3 \)[/tex] to [tex]\( \text{dm}^3 \)[/tex]: [tex]\( 25 \, \text{cm}^3 \times \frac{1 \, \text{dm}^3}{1000 \, \text{cm}^3} = 0.025 \, \text{dm}^3 \)[/tex]
- Concentration of NaOH solution: [tex]\( 0.8 \, \text{mol} \cdot \text{dm}^{-3} \)[/tex]
2. Calculate the Moles of NaOH:
- Using the formula [tex]\( \text{moles} = \text{volume} \times \text{concentration} \)[/tex]:
[tex]\[ \text{moles of NaOH} = 0.025 \, \text{dm}^3 \times 0.8 \, \text{mol} \cdot \text{dm}^{-3} = 0.020 \, \text{mol} \][/tex]
3. Determining the Moles of Excess HCl:
- Since NaOH reacts with HCl in a 1:1 molar ratio:
[tex]\[ \text{moles of excess HCl} = \text{moles of NaOH} = 0.020 \, \text{mol} \][/tex]
4. Initial Moles of HCl:
- Volume of HCl solution: [tex]\( 100 \, \text{cm}^3 \)[/tex]
- Convert volume from [tex]\( \text{cm}^3 \)[/tex] to [tex]\( \text{dm}^3 \)[/tex]: [tex]\( 100 \, \text{cm}^3 \times \frac{1 \, \text{dm}^3}{1000 \, \text{cm}^3} = 0.1 \, \text{dm}^3 \)[/tex]
- The solution is in excess, and the neutralized amount is the excess moles:
[tex]\[ \text{initial moles of HCl} = \text{moles of excess HCl} + \text{moles of NaOH} \][/tex]
[tex]\[ \text{initial moles of HCl} = 0.020 \, \text{mol} + 0.020 \, \text{mol} = 0.040 \, \text{mol} \][/tex]
5. HCl That Reacted with [tex]\( MCO_3 \)[/tex]:
- Moles of HCl that reacted with [tex]\( MCO_3 \)[/tex]:
[tex]\[ \text{moles of HCl reacted} = \text{initial moles of HCl} - \text{moles of excess HCl} = 0.040 \, \text{mol} - 0.020 \, \text{mol} = 0.020 \, \text{mol} \][/tex]
6. Moles of [tex]\( MCO_3 \)[/tex]:
- Based on the balanced chemical equation:
[tex]\[ MCO_3 + 2HCl \rightarrow \text{products} \][/tex]
- It takes 2 moles of HCl to react with 1 mole of [tex]\( MCO_3 \)[/tex]:
[tex]\[ \text{moles of } MCO_3 = \frac{\text{moles of HCl reacted}}{2} = \frac{0.020 \, \text{mol}}{2} = 0.010 \, \text{mol} \][/tex]
7. Finding the Molar Mass of [tex]\( MCO_3 \)[/tex]:
- Given mass of [tex]\( MCO_3 \)[/tex]: [tex]\( 3.5 \, \text{g} \)[/tex]
- Using the formula [tex]\( \text{molar mass} = \frac{\text{mass}}{\text{moles}} \)[/tex]:
[tex]\[ \text{molar mass of } MCO_3 = \frac{3.5 \, \text{g}}{0.010 \, \text{mol}} = 350 \, \text{g} \cdot \text{mol}^{-1} \][/tex]
8. Finding the Molar Mass of the Metal [tex]\( M \)[/tex]:
- Molar mass of [tex]\( CO_3 \)[/tex]: [tex]\( 12 \, (\text{C}) + 16 \times 3 \, (\text{O}) = 60 \, \text{g} \cdot \text{mol}^{-1} \)[/tex]
- Molar mass of [tex]\( MCO_3 \)[/tex]: [tex]\( 350 \, \text{g} \cdot \text{mol}^{-1} \)[/tex]
- Molar mass of metal [tex]\( M \)[/tex] is found by subtracting molar mass of [tex]\( CO_3 \)[/tex] from molar mass of [tex]\( MCO_3 \)[/tex]:
[tex]\[ \text{Molar mass of } M = 350 \, \text{g} \cdot \text{mol}^{-1} - 60 \, \text{g} \cdot \text{mol}^{-1} = 290 \, \text{g} \cdot \text{mol}^{-1} \][/tex]
### Conclusion:
The molar mass of the unknown metal [tex]\( M \)[/tex] is [tex]\( 290 \, \text{g} \cdot \text{mol}^{-1} \)[/tex]. Given this specific molar mass, the metal is likely to be Rhenium ([tex]\( \text{Re} \)[/tex]), which has a molar mass close to 290 g/mol.
Therefore, the symbol of the unknown metal [tex]\( M \)[/tex] is [tex]\( \text{Re} \)[/tex].
### Step-by-Step Calculation:
1. Volume and Concentration of Sodium Hydroxide Solution:
- Volume of NaOH solution: [tex]\( 25 \, \text{cm}^3 \)[/tex]
- Convert volume from [tex]\( \text{cm}^3 \)[/tex] to [tex]\( \text{dm}^3 \)[/tex]: [tex]\( 25 \, \text{cm}^3 \times \frac{1 \, \text{dm}^3}{1000 \, \text{cm}^3} = 0.025 \, \text{dm}^3 \)[/tex]
- Concentration of NaOH solution: [tex]\( 0.8 \, \text{mol} \cdot \text{dm}^{-3} \)[/tex]
2. Calculate the Moles of NaOH:
- Using the formula [tex]\( \text{moles} = \text{volume} \times \text{concentration} \)[/tex]:
[tex]\[ \text{moles of NaOH} = 0.025 \, \text{dm}^3 \times 0.8 \, \text{mol} \cdot \text{dm}^{-3} = 0.020 \, \text{mol} \][/tex]
3. Determining the Moles of Excess HCl:
- Since NaOH reacts with HCl in a 1:1 molar ratio:
[tex]\[ \text{moles of excess HCl} = \text{moles of NaOH} = 0.020 \, \text{mol} \][/tex]
4. Initial Moles of HCl:
- Volume of HCl solution: [tex]\( 100 \, \text{cm}^3 \)[/tex]
- Convert volume from [tex]\( \text{cm}^3 \)[/tex] to [tex]\( \text{dm}^3 \)[/tex]: [tex]\( 100 \, \text{cm}^3 \times \frac{1 \, \text{dm}^3}{1000 \, \text{cm}^3} = 0.1 \, \text{dm}^3 \)[/tex]
- The solution is in excess, and the neutralized amount is the excess moles:
[tex]\[ \text{initial moles of HCl} = \text{moles of excess HCl} + \text{moles of NaOH} \][/tex]
[tex]\[ \text{initial moles of HCl} = 0.020 \, \text{mol} + 0.020 \, \text{mol} = 0.040 \, \text{mol} \][/tex]
5. HCl That Reacted with [tex]\( MCO_3 \)[/tex]:
- Moles of HCl that reacted with [tex]\( MCO_3 \)[/tex]:
[tex]\[ \text{moles of HCl reacted} = \text{initial moles of HCl} - \text{moles of excess HCl} = 0.040 \, \text{mol} - 0.020 \, \text{mol} = 0.020 \, \text{mol} \][/tex]
6. Moles of [tex]\( MCO_3 \)[/tex]:
- Based on the balanced chemical equation:
[tex]\[ MCO_3 + 2HCl \rightarrow \text{products} \][/tex]
- It takes 2 moles of HCl to react with 1 mole of [tex]\( MCO_3 \)[/tex]:
[tex]\[ \text{moles of } MCO_3 = \frac{\text{moles of HCl reacted}}{2} = \frac{0.020 \, \text{mol}}{2} = 0.010 \, \text{mol} \][/tex]
7. Finding the Molar Mass of [tex]\( MCO_3 \)[/tex]:
- Given mass of [tex]\( MCO_3 \)[/tex]: [tex]\( 3.5 \, \text{g} \)[/tex]
- Using the formula [tex]\( \text{molar mass} = \frac{\text{mass}}{\text{moles}} \)[/tex]:
[tex]\[ \text{molar mass of } MCO_3 = \frac{3.5 \, \text{g}}{0.010 \, \text{mol}} = 350 \, \text{g} \cdot \text{mol}^{-1} \][/tex]
8. Finding the Molar Mass of the Metal [tex]\( M \)[/tex]:
- Molar mass of [tex]\( CO_3 \)[/tex]: [tex]\( 12 \, (\text{C}) + 16 \times 3 \, (\text{O}) = 60 \, \text{g} \cdot \text{mol}^{-1} \)[/tex]
- Molar mass of [tex]\( MCO_3 \)[/tex]: [tex]\( 350 \, \text{g} \cdot \text{mol}^{-1} \)[/tex]
- Molar mass of metal [tex]\( M \)[/tex] is found by subtracting molar mass of [tex]\( CO_3 \)[/tex] from molar mass of [tex]\( MCO_3 \)[/tex]:
[tex]\[ \text{Molar mass of } M = 350 \, \text{g} \cdot \text{mol}^{-1} - 60 \, \text{g} \cdot \text{mol}^{-1} = 290 \, \text{g} \cdot \text{mol}^{-1} \][/tex]
### Conclusion:
The molar mass of the unknown metal [tex]\( M \)[/tex] is [tex]\( 290 \, \text{g} \cdot \text{mol}^{-1} \)[/tex]. Given this specific molar mass, the metal is likely to be Rhenium ([tex]\( \text{Re} \)[/tex]), which has a molar mass close to 290 g/mol.
Therefore, the symbol of the unknown metal [tex]\( M \)[/tex] is [tex]\( \text{Re} \)[/tex].
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