Answer:
0.18 moles
Explanation:
Applying,
PV = nRT................... Equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (0.8316×5.3)/(0.083×295)
n = 0.18 moles
Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.
Explanation :
First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.
As, 207.2 g of lead contains
atoms
So, 5.5 g of lead contains
atoms
and,
As, 118.71 g of lead contains
atoms
So, 94.5 g of lead contains
atoms
Now we have to calculate the percent composition of Pb and Sn in atom.


and,


Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.
Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have
as nucleophile. Also, this compound is also in excess. So, we will have as solvent
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).