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ppb of chcl3 in drinking water means 100 g of CHCl3 in 1,000,0000,000 g of water
Molarity, M
M = number of moles of solute / volume of solution in liters
number of moles of solute = mass of CHCl3 / molar mass of CHCl3
molar mass of CHCl3 = 119.37 g/mol
number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol
using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters
M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer
Molality, m
m = number of moles of solute / kg of solvent
number of moles of solute = 0.838
kg of solvent = kg of water = 1,000,000 kg
m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer
mole fraction of solute, X solute
X solute = number of moles of solute / number of moles of solution
number of moles of solute = 0.838
number of moles of solution = number of moles of solute + number of moles of solvent
number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles
number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles
X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer
mass percent, %
% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =
% = 10 ^ - 6 % <------- answer
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The organic compound retinal binds with opsin and forms rhodopsin. Retinal is part of the molecule that is responsible for its color. This part is called chromophore. On the other hand, opsins are the proteins in photoreceptor cells. Retinal bounds with these opsins and forms rhodopsin: the basis of the human vision. Rhodopsin is also a protein.It is the pigment in the retinas of humans and animals.
Answer:
(A) The work done by the system is -101.325J
(B) The workdone by the system is -90.75J
Explanation:
(A) Workdone = -PΔV
Given that A = 100cm2 = 0.01m2
distance d = 10cm = 0.1m
ΔV= Area × distance
ΔV= 0.01 ×0.1
ΔV = 0.001m3
P= external pressure = 1atm = 101325Pa
Workdone = -0.001 × 101325
W= - 101.325Pa m3
1Pam3 = 1J
Therefore W = - 101.325J
The work done on the system is -101.325J
(B) Workdone = -PΔV
Given that A = 50cm2 = 0.005m2
distance d = 15cm = 0.15m
ΔV= Area × distance
ΔV= 0.005×0.15
ΔV = 0.00075m3
P=121kPa = 121000Pa
W= - 121000 × 0.00075
W= -90.75Pa m3
1Pam3 = 1J
W = - 90.75J
The woekdone by the system is -90.75J
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