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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Answer:
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Mass of molecule (g) = Mr of substance over avarogado constant
Answer:
1. B
2. M
3. V
4. M
5. M
6. V
7. B
8. M
9. V
10. V
11. Red
12. Approx. 1/4 - 1/2 an earth year
Explanation: