4.22 grams.
1. First find out how much AgNO3 weighs with one mole (107.87 g Ag + 14.007 g N + 48 g O = 169.89 grams)
2. Find the percent of Ag you have. So, (107.87 g/mol Ag)/(169.89 g/mol AgNO3)= 0.63 * 100 = 63%.
3. If you have 6.7 grams total, you know 63% of it is going to be silver, so just multiply 6.7 grams by .63 and you get 4.22 g Ag
Answer: solvent
solvent is A solvent is usually a liquid
2.47 moles are produce if 85g of N2 react with 180g of Mg
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:
