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:
H2(g) + Cl2(g) = 2 HCl(aq) (balanced equation)
1, 1, 2 (coefficients)
Answer:
The steps for solving a numeric word problem are analyze, calculate, and evaluate. To solve a word problem, you must first determine where you are starting from (identify what is known) and where you are going (identify the unknown).
Answer: 1.348 ×10^23 atoms
Explanation:
Given that volume = 1.00L
At standard condition, the volume of a gas is 22.4L/mol (at S.T.P)
Volume = mole /volume at STP
1 = mole/22.4
Mole= 22.4mol.
Also
Mole = number of atoms /Avogradro constant
Where avogrado's constant = 6.02×10²³
22.4 = number of atoms/6.02×10²³
Number of atoms = 1.348×10^25atoms
1. 254 cal = 1,062.736 joules
2. 126 cal = 527.184 joules
3. 98 cal = 410.032 joules
4. 704 cal = 2,945.536 joules
5. 682 cal = 2,853.488 joules
