Answer:- 13.6 L
Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.
Standard pressure is 1 atm that is 101.325 Kpa.
Boyle's law equation is:
From given information:-
= 58.7 Kpa
= 23.5 L
= 101.325 Kpa
= ?
Let's plug in the values and solve it for final volume.
On rearranging the equation for
= 13.6 L
So, the volume of hydrogen gas at STP for the given information is 13.6 L.
Answer:
28 is great.......... it should be snSwer
Cu⇒ 1 atom
N⇒2 atoms
O⇒6 atoms
Total 9 atoms
<h3>Further explanation </h3>
The empirical formula is the smallest comparison of atoms of compound forming elements.
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
<em>(empirical formula) n = molecular formula </em>
Chemical formula : Cu(NO₃)₂
Number of Cu : 1
Number of N : 2
Number of O = 2 x 3 = 6
Total atoms in Cu(NO₃)₂ : 1 + 2 + 6 = 9
1.4715 atm is the pressure of the sample 1.33 moles of fluorine gas that is contained in a 23.3 L container at 314 K.
What is an ideal equation?
The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa).
Given data:
Volume (V) = 23.3 L
Number of mole (n) = 1.33 moles
Temperature (T) = 314 K
Gas constant (R) = 0.821 atm.L/Kmol
Pressure (P) =?
The pressure inside the container can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
P × 23.3 L = 1.33 moles × 0.0821 ×314 K
P = 1.4715 atm
Therefore, the pressure of the sample is 1.4715 atm.
Learn more about the ideal gas equation:
brainly.com/question/23826793
SPJ1
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:
