Answer:
all good. tell me about your day
Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of
b) moles of
According to stoichiometry :
1 mole of
require 1 mole of
Thus 0.0787 moles of
require=
of
Thus
is the limiting reagent as it limits the formation of product and
acts as the excess reagent. (10.0-0.0787)= 9.92 moles of
are left unreacted.
Mass of
Thus 19.9 g of
remains unreacted.
Answer:
y1 = 0.3162
y2 = 0.6838
Explanation:
ok let us begin,
first we would be defining the parameters;
at 25°C;
1-propanol P1° = 20.90 Torr
2-propanol P2° = 45.2 Torr
From Raoults law:
P(1-propanol) = P⁰ × X(1-propanol)
P(1-propanol) = 20.9 torr × 0.45 = 9.405
P(1-propanol) = 9.405 torr
Also P(2-propanol) = P⁰ × X(2-propanol)
P(2-propanol) = 45.2 torr × 0.45
P(2-propanol) = 20.34 torr
but the total pressure = sum of individual pressures
total pressure = 9.405 + 20.34
total pressure = 29.745 torr
given that y1 and y2 represent the mole fraction of each in the vapor phase
y1 = P1 / total pressure
y1 = 9.405/29.745
y1 = 0.3162
Since y1 + y2 = 1
y2 = 1 - y1
∴ y2 = 1 - 0.3162
y2 = 0.6838
cheers, i hope this helps.
Answer:
Cr (HSO4)3
Explanation:
its molecular weight is 343.20 g/mol
its molecular formula can also be written as CrH3O12S3
molar mass of Cr (HSO4)3 can be calculated by following method;
atomic mass of Cr = 51.9961 u
atomic mass of H = 1 u
atomic mass of S = 32.065 u
atomic mass of O = 16 u
molar mass of Cr(HSO4)3 = 51.9961+ 1.00784×3 + 32.065×3 + 15.999×12
molar mass of Cr(HSO4)3 =51.9961+3.02352+96.195+ 191.988
molar mass of Cr(HSO4)3 = 343.20 g/mol
<h2>Natural Abundance for 10B is 19.60%</h2>
Explanation:
- The natural isotopic abundance of 10B is 19.60%.
- The natural isotopic abundance of 11B is 80.40%.
- The isotopic masses of boron are 10.0129 u and 11.009 u respectively.
For calculation of abundance of both the isotopes -
Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.
Determining it as an equation -
10x + 11y= 10.8
x+y=1 (ratio)
10x + 10y = 10
By taking the denominator away from the numerator
we get;
y = 0.8
x + y = 1
∴ x = 0.2
To get percentages we need to multiply it by 100
So, the calculated abundance is 80% for 11 B and 20% 10 B.