Answer:
1. Percentage by weight = 0.5023 = 50.23 %
2. molar fraction =0.153
Explanation:
We know that
Molar mass of HClO4 = 100.46 g/mol
So the mass of 5 Moles= 5 x 100.46
Mass (m)= 5 x 100.46 = 502.3 g
Lets assume that aqueous solution of HClO4 and the density of solution is equal to density of water.
Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.
We know that
Mass = density x volume
Mass of 1000 ml solution = 1 x 1000 =1000 ( density = 1 gm/ml)
m'=1000 g
1.
Percentage by weight = 502.3 /1000
Percentage by weight = 0.5023 = 50.23 %
2.
We know that
molar mass of water = 18 g/mol
mass of water in 1000 ml = 1000 - 502.3 g=497.9 g
So moles of water = 497.7 /18 mole
moles of water = 27.65 moles
So molar fraction = 5/(5+27.65)
molar fraction =0.153
DDT was one of the first chemicals in widespread use as a pesticide.
<span><span>Food supplies: USDA found DDT breakdown products in 60% of heavy cream samples, 42% of kale greens, 28% of carrots and lower percentages of many other foods.</span><span>Body burden: DDT breakdown products were found in the blood of 99% of the people tested by CDC.</span><span>Health impacts: Girls exposed to DDT before puberty are 5 times more likely to develop breast cancer in middle age, according to the President’s Cancer Panel.</span></span>
Answer:
increase
Explanation:
At the skate park when a skateboarder wants to skate down a ramp, the skate boarder wishes to move faster and with speed so that it help him to skate more time. This can only be achieved if the all the forces acting on him will increase as he skates down the ramp without much of the frictional force acting on the wheels of the skate board. The less friction force acts on the wheel, the more he can skate with more speed.
Thus, the skateboarder wants the force to be increase that is acting on him.
It required a fixed finite amount these zones are known as energy levels
Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
