Answer:
Partial pressure of oxygen = 23.38 kpa (Approx)
Explanation:
Given:
Amount of oxygen = 23.15%
Amount of nitrogen = 76.85%
Pressure (missing) = 101 kpa
Find:
Partial pressure of oxygen
Computation:
Partial pressure of oxygen = [Amount of oxygen x Pressure]/100
Partial pressure of oxygen = [23.15% x 101]/100
Partial pressure of oxygen = 23.38 kpa (Approx)
(3 x 10^10 Hg) x (1 mole Hg/6.022 x 10^23 Hg) x (200.59g / 1 mole Hg) =
99.928 x 10^-13
10^-9 g Hg x (1 mole Hg / 200.59 g Hg) x (6.022 x 10^23 Hg / 1 mole Hg) =
0.031 x 10^14 Hg = 3.1 x 10^12 Hg
I.) 10^-11 g Hg
II.) 3.1 x 10^12 Hg
Answer:
2.01
Explanation:
First, let's convert grams to moles
(Na) 22.99 + (F) 18.998 = 41.988
Every mole of NaF is 41.988 grams
21/41.988 = 0.500143 moles of NaF
For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF
0.500143/2 = 0.250071
molarity = moles/liters
0.250071/0.125 = 2.0057 M
0.05 is the answer to your question
