1 mole of platinum has a mass of 195 g therefore 1 atom will have a mass of 195 g /(6.02 ×10^23) = 3.239 × 10^-22 g
Density is given by dividing mass by volume, thus to get volume, mass is divided by density.
The volume = (3.239 × 10^-22)/21.4
= 1.514 × 10^-23 cm³
But volume of a sphere is given by 4/3πr³
Therefore, r³ = 3.6129 × 10^-24
r = ∛(3.6129 × 10^-24)
= 1.534 × 10^ -8 cm
Therefore, the radius of the platinum atom is 1.534 × 10^-8 cm
Answer:
C. 1.35
Explanation:
2NH3 (g) <--> N2 (g) + 3H2 (g)
Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0
change in concentration 2x x 3x
-0.84 M +0.42M +1.26M
Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M
concentration
Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M
Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M
Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M
K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M
Answer:
N₂ = 0.7515atm
O₂ = 0.1715atm
NO = 0.0770atm
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO(g)
Where Kp is defined as:
Pressures in equilibrium are:
N₂ = 0.790atm - X
O₂ = 0.210atm - X
NO = 2X
Replacing in Kp:
0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]
0.0460 = 4X² / 0.1659 - X + X²
0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²
-3.954X² - 0.0460X + 7.6314x10⁻³ = 0
Solving for X:
X = - 0.050 → False answer. There is no negative concentrations.
X = <em>0.0385 atm</em> → Right answer.
Replacing for pressures in equilibrium:
N₂ = 0.790atm - X = <em>0.7515atm</em>
O₂ = 0.210atm - X = <em>0.1715atm</em>
NO = 2X = <em>0.0770atm</em>