Answer:
40.5% is abundance of X-122
Explanation:
we know there are two naturally occurring isotopes of X, X-120 and X-122.
First of all we will set the fraction for both isotopes
X for the isotopes having mass 122
1-x for isotopes having mass 12
0
The average atomic mass of X is 120.81 amu
we will use the following equation,
122x + 120 (1-x) = 120.81
122x + 120 - 120x = 120.81
122x- 120x = 120.81 -120
2x = 0.81
x= 0.81/2
x= 0.405
0.405 × 100 = 40.5%
40.5% is abundance of X-122 because we solve the fraction x.
now we will calculate the abundance of X-120.
(1-x)
1 - 0.405 = 0.595
0.595 × 100 = 59.5 %
59.5 % for X-120.
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)
R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide
Answer:
Metalloids are semiconductive.
Explanation:
Answer:
A-20 B-40 C-Ca D-10 E-9 F-F