The mass of sodium chloride at the two parts are mathematically given as
- m=10,688.18g
- mass of Nacl(m)=39.15g
<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>
Generally, the equation for ideal gas is mathematically given as
PV=nRT
Where the chemical equation is
F2 + 2NaCl → Cl2 + 2NaF
Therefore
1.50x15=m/M *(1.50*0.0821)
1-50 x 15=m/58.5 *(1.50*0.0821)
m=10,688.18g
Part 2
PV=m'/MRT
1*15=m'/58.5*0.0821*273
m'=39.15g
mass of Nacl(m)=m'=39.15g
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Propylethylene would be the answer
CaBr2(s) Ca+2(aq)+2Br-(aq) which means, <span>Solid is turning into free ions so entropy is increasing .</span>
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.
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