The balanced chemical reaction is written as:
<span>CH4 (g) + 2 O2 (g) ----> CO2 (g) + 2 H2O (g)
</span>
We are given the amount of water to be produced from the reaction. This amount will be used for the calculations. Calculations are as follows:
12.4 L H2O ( 1 mol / 22.4 L ) ( 1 mol CH4 / 2 mol H2O ) ( 22.4 L / 1 mol ) = 6.2 L CH4
Answer:
70.15 cm³
Solution:
Data Given;
Mass = 55 g
Density = 0.784 g.cm⁻³
Required:
Volume = ?
Formula Used:
Density = Mass ÷ Volume
Solving for Volume,
Volume = Mass ÷ Density
Putting values,
Volume = 55 g ÷ 0.784 g.cm⁻³
Volume = 70.15 cm³
Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.
Explanation :
As we know that:
At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.
The relation between pressure and number of moles of gas will be:
where,
= initial pressure of gas = 24.5 atm
= final pressure of gas = 5.30 atm
= initial number of moles of gas = 1.40 moles
= final number of moles of gas = ?
Now put all the given values in the above expression, we get:
Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.
The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
Learn more about isotope:
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Answer:
467
Explanation:
ncl2 = 454.4x1/(71.0 g/mol) = 6.40 mols cl2
6.40 mols cl2 x 2molsHCL/1moleCL2 x 36.5g/1moleHCL = <u>467 g HCL</u>
