Answer:
The answer to your question is 80.3%
Explanation:
Data
Percent by mass of F
molecules NF₃
Process
1.- Calculate the molar mass of nitrogen trifluoride
molar mass = (1 x 14) + (19 x 3)
= 14 + 57
= 71 g
2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.
71 g of NF₃ ------------------ 100%
57 g of F ------------------- x
x = (57 x 100)/71
x = 5700 / 71
x = 80.3%
3.- Conclusion
Fluorine is 80.3% by mass of the molecule NF₃
Answer:
0.846 moles.
Explanation:
- This is a stichiometric problem.
- The balanced equation of complete combustion of butane is:
C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O
- It is clear from the stichiometry of the balanced equation that complete combustion of 1.0 mole of butane needs 6.5 moles of O₂ to produce 4 moles of CO₂ and 5 moles of H₂O.
<u><em>Using cross multiplication:</em></u>
- 1.0 mole of C₄H₁₀ reacts with → 6.5 moles of O₂
- ??? moles of C₄H₁₀ are needed to react with → 5.5 moles of O₂
- The number of moles of C₄H₁₀ that are needed to react with 5.5 moles of O₂ = (1.0 x 5.5 moles of O₂) / (6.5 moles of O₂) = 0.846 moles.
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.