Explanation:
a) In 1 mole of methane there are 4 moles of hydrogen atom
Atomic mass of 1 mole of hydrogen atom = 1 g
Mass of hydrogen in 1 mole of methane = 4 × 1 g = 4 g
b) In 1 mole of chloroform there are 1 mole of hydrogen atom
Atomic mass of 1 mole of hydrogen atom = 1 g
Mass of hydrogen in 1 mole of methane = 1× 1 g = 1 g
c) In 1 mole of
there are 10 moles of hydrogen atom
Atomic mass of 1 mole of hydrogen atom = 1 g
Mass of hydrogen in 1 mole of
= 10 × 1 g = 10 g
d)In 1 mole of
there are 12 moles of hydrogen atom.

Atomic mass of 1 mole of hydrogen atom = 1 g
Mass of hydrogen in 1 mole of
= 12 × 1 g = 12 g
Answer:
72.53% is the yield of CrCl3
Explanation:
Given
Reaction:
Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)
CCl4 is in excess and 17.6g Cr2O3 present
The reaction yields 26.6g of CrCl3
To Find:
% yields of the reaction
Also given
Molar mass of CrCl3 = 158.35g/mol
Molar mass of Cr2O3 = 152.00 g/mol
By the stoichiometry of the reaction
1 mole of Cr2O3 gives 2 moles of CrCl3
0r
1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3
= 1 52 g of Cr2O3 gives 316.70 g of CrCl3
17.6 g of Cr2O3 gives (17.6÷152) × 316.70 g CrCl3
= 36.67 g CrCl3
but actual yield is only 26.6g
so % yield is (26.6 ÷÷ 36.67) × 100
= 72.53% is the yield of CrCl3
Ideal gases are hypothetical gases whose molecules occupy negligible space and have no interactions, and that consequently obeys the gas laws exactly.
Not exactly sure about the amount...
I hope this helps! :)
The scientist's results is that at a temperature of 35<span>°C, the solubility of the substance in water is 146.2 grams in 200 grams of water. There isn't really a different method to determine the solubility of a substance in water. Another procedure could be that a lesser amount of the substance is used and the water required to dissolve it is determined. The solubility of the substance based on the two procedures can then be compared.</span>
Answer:
D. 1.48atm
Explanation:
Van der waals equation is given as:
(P +an²/v²) (v - nb) = nRT
Where;
P = pressure (atm)
V = volume (L)
R = gas constant (0.0821 Latm/molK)
a and b = gas constant specific to each gas
T = temperature (K)
n = number of moles
According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol
Hence;
(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15
(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638
(P + 0.0291) (22.316) = 33.638
22.316P + 0.649 = 33.638
22.316P = 33.638 - 0.649
22.316P = 32.989
P = 32.989/22.316
P = 1.478
P = 1.48atm