2.4 Bromium has two naturally occurring isotopes: 79Br, with an atomic weight of 78.918 amu, and 81Br, with an atomic weight of
1 answer:
Answer:
49.3% 81Br and 50.7% 79Br or 81Br and
79BR
Step-by-step explanation:
To get the fraction-of-occurrences of the isotopes we must write the following equation. x is the isotopic abundance of 81Br, we can use 1 - x to get the isotopic abundance of 79Br.
(78.918)(1 - x) + (80.916)(x) = 79.903
78.918 - 78.918x + 80.916x = 79.903
1.998x = 0.985
x = 0.493
0.493 × 100 = 49.3% 81Br
100 - 49.3 = 50.7% 79Br
49.3% 81Br and 50.7% 79Br or 81Br and
79BR
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Answer:
1223.38 mmHg
Explanation:
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value =
Also,
Moles = mass (m) / Molar mass (M)
Density (d) = Mass (m) / Volume (V)
So, the ideal gas equation can be written as:
Given that:-
d = 1.80 g/L
Temperature = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (32 + 273.15) K = 305.15 K
Molar mass of nitrogen gas = 28 g/mol
Applying the equation as:
P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K
⇒P = 1223.38 mmHg
<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>