Answer:
The mean free path = 2.16*10^-6 m
Explanation:
<u>Given:</u>
Pressure of gas P = 100 kPa
Temperature T = 300 K
collision cross section, σ = 2.0*10^-20 m2
Boltzmann constant, k = 1.38*10^-23 J/K
<u>To determine:</u>
The mean free path, λ
<u>Calculation:</u>
The mean free path is related to the collision cross section by the following equation:
where n = number density
Substituting for P, k and T in equation (2) gives:
Next, substituting for n and σ in equation (1) gives:
Hey there!
<span>Use the equation of Clapeyron:
</span>
T in kelvin :
26 + 273.15 => 299.15 K
R = 0.082
V = 10.2 L
P = 0.98 atm
number of moles :
P *V = n * R * T
0.98 * 10.2 = n * 0.082 * 299.15
9.996 = n * 24.5303
n = 9.996 / 24.5303
n = 0.4074 moles
Therefore:
Molar mass H2O = 18.01 g/mol
1 mole H2O ------------- 18.01 g
0.4074 moles ----------- m
m = 0.4074 * 18.01 / 1
m = 7.339 g of H2O
From the given chemical formula or equation, it can be seen that for every molecule of the given chemical, there are 2 moles of NH4+ ions. Thus, to calculate for the concentration of the ion in the solution, we multiply 0.30M by 2 giving as an answer of 0.60M.
Answer:
193.07 g / mol
Explanation:
Molar mass of (NH₄)₃ = 3 * (14.01 + 4 * 1.01) = 54.15
As = 74.92 and O₄ = 4 * 16 = 64
Answer is 54.15 + 74.92 + 64 = 193.07
Answer:
10.13086%
Explanation:
129y + 132(1-y) = 129.91
129y + 132 - 132y = 129.91
-3y = -2.09
y = 0.69666 (5 s.f.)
% abundance of 132X = 100% - (129×0.69666)%
= 100% - 89.86914%
= 10.13086%
= 10.1% (3 significant figures)
*y is just a variable for the equation
