Answer:
Three times with 5 ml will yield more
Explanation:
Let x represent the amount yield
Kd = (x/15) / ((50-x) / 15) where Kd = 1.5
1.5 = (x/15) / ((50-x) / 15)
x / (50 - x) = 1.5
x = 75 - 1.5x
x + 1.5x = 75
2.5x = 75
x = 75 / 2.5 = 30 mg
when extraction three times
1st extraction
(x1/5) / ((50 - x1) / 15) = 1.5
3x1 / 50 - x1 = 1.5
3x1 = 75 - 1.5x1
3x1 + 1.5x1 = 75
4.5x1 = 75
x1 = 75 / 4.5 = 16.67 gm
second extraction
(x2/ 5) / (33.33 - x2 ) / 15) = 1.5
3x2 / ( 33.33 - x2) = 1.5
3x2 = 1.5(33.33 - x2)
3x2 = 49.995 - 1.5x2
3x2 + 1.5x2 = 49.995
4.5x2 = 49.995
x2 = 49.995 / 4.5
x2 = 11.11 mg
Third extraction
(x3/5) / ((22.22 - x3) / 15) = 1.5
3x3 = 1.5 ( 22.22 - x3)
3x3 + 1.5x3 = 33.33
4.5x3 = 33.33
x3 = 33.33 / 4.5 = 7.41 mg
total extraction = x1 + x2 + x3 =16.67 + 11.11 + 7.41 = 35.19 mg
the three times extraction using 5ml yields 5.19 mg more
33.6 moles are needed to completely react with 84.0 moles of O2
Explanation:
4 x 6.02 x 10²³ = 2.41 x 10²⁴
The number of molecules that are in balloon are = 2.227 x10^23 molecules
<h3> calculation</h3>
calculate the number of moles of NO
moles = mass/molar mass
molar mass of NO = 14+ 16 = 30 g/mol
moles is therefore= 11.1 g/30g/mol= 0.37 moles
by use of Avogadro's constant that is
1 mole= 6.02 x10^23 molecules
0.37 =? molecules
=(6.02 x10^23 x 0.37 moles)/ 1mole=2.227 x10^23 molecules
Answer:
Nitrogen
Oxygen
Argon
Carbon Dioxide
Methane
Ozone
Explanation:
N₂ accounts for 78% of the atmosphere.
O₂ accounts for 21% of the atmosphere.
Ar accounts for 0.9% of the atmosphere.
CO₂, CH₄, and O₃ only take up 0.1% of the atmosphere.
