On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
Answer:
The empirical formula of compound is C₂H₆O.
Explanation:
Given data:
Mass of carbon = 12 g
Mass of hydrogen = 3 g
Mass of oxygen = 8 g
Empirical formula of compound = ?
Solution:
First of all we will calculate the gram atom of each elements.
no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms
no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms
no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms
Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.
C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5
C:H:O = 2 : 6 : 1
The empirical formula of compound will be C₂H₆O
Answer:
b
. Irradiated food is shown to not be radioactive.
Explanation:
If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.
- An irradiated food is one in which ionizing radiations have been employed to improve food quality.
- Thus, bacteria and other food spoilers can be exterminated from the food.
- Most irradiated food do not contain radiation and are fit for consumption.
If it can be proven, that this is true, then it will challenge the idea that irradiated foods are not safe.
Answer:
See explanation
Explanation:
% optical purity = specific rotation of mixture/specific rotation of pure enantiomer * 100/1
specific rotation of mixture = 23°
specific rotation of pure enantiomer = 61°
Hence;
% optical purity = 23/61 * 100 = 38 %
More abundant enantiomer = 100% - 38 % = 62%
Hence the pure (S) carvone is (-) 62° is the more abundant enantiomer.
Enantiomeric excess = 62 - 50/50 * 100 = 24%
Hence
(R) - carvone = 38 %
(S) - carvone = 62%
Convert grams —> mols and then mols —> atoms
We know that there are 6.02 x 10^23 atoms/mol
And we know that there are about 160 grams of fe2o3 per mol
So (79g fe2o3)/(160 g/mol) = .49 mol fe2o3
Now we use avogadro’s number to do
(.49 mol fe2o3)/(6.02 x 10^23 atoms/mol) = the answer.
I’ll leave the easy math to you.
