Answer:
61 mole % propylcyclohexane and 39 mole % propylbenzene
Explanation:
For convenience, let’s call propylcyclohexane <em>Component 1 </em>and propylbenzene <em>Component 2</em>.
According to <em>Raoult’s Law</em>,
and
where
<em>p</em>₁ and <em>p</em>₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole \fractions of the components
<em>p</em>₁° and <em>p</em>₂° are the vapour pressures of the pure components.
So,
= <em>p</em>₁ + <em>p</em>₂= 461 torr + 290 torr = 751 torr
∴ In the vapour
χ₂ = 1 – χ₁ = 1 - 0.61 = 0.39
Answer:
Mass= 2.77g
Explanation:
Applying
P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28
PV=nRT
NB
Moles(n) = m/M
PV=m/M×RT
m= PVM/RT
Substitute and Simplify
m= (2.09×1.13×28)/(0.082×291)
m= 2.77g
Answer: they are the same.
Explanation:
1) Sucrose is a compound with chemical formula C₁₂H₂₂O₁₁
2) That means that all molcules of sucrose will have the same kind of atoms in the same proportion, whic is to say same composition:
12 atoms of C: 22 atoms of H: 11 atoms of O, per each molecule of sucrose.
3) For this question you can rely in the definition of compound: a pure substance formed by the combination of two or more elements always in the same ratio (same composition).
4) That also implies, that all the molecules of sucrose have the same properties.
Answer: The correct answer is: the water can transfer heat to your arm more quickly than the air.
Explanation: The heat is transferred from the air or water to your arm through convection. The convective heat transfer coefficient of water is higher than the air's, so, even though the temperature of boiling water is lower, the heat will be transferred more efficiently to the other surface, in this case, the hand.
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
