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.
B .pure water it’s made up of oxygen and hydrogen “H2O”
Answer: In the first paragraph, name a theme of Paul Laurence Dunbar's poem "Sympathy," and explain how it develops, citing specific examples
Explanation:
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
A rock is definitely more dense. If you were to put a cloud in water it would float/stay above it and a rock would sink to the bottom
