Answer:
His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.
Explanation:
The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.
Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.
Suppose the mass of the solute is m.
Originally, the density is = 
Now after adding extra 10 mL , the density becomes 
.
Therefore, 
So the density decreases when we add more solution.
Answer:
29260J
Explanation:
Given parameters:
Mass of water sample = 100g
Initial temperature = 30°C
Final temperature = 100°C
Unknown:
Energy required for the temperature change = ?
Solution:
The amount of heat required for this temperature change can be derived from the expression below;
H = m c (ΔT)
H is the amount of heat energy
m is the mass
c is the specific heat capacity of water = 4.18J/g°C
ΔT is the change in temperature
Now insert the parameters and solve;
H = 100 x 4.18 x (100 - 30)
H = 100 x 4.18 x 70 = 29260J
I am just answering so u can mark the other guy brainliest
Answer: mole fractions are
For n-pentane = 0.965
For n-octane = 0.035
Explanation: pressure exerted by each gas is,
n-pentane = 1.92atm
n-octane = 0.07atm
Total pressure exerted = 1.92 + 0.07
= 1.99atm.
Recall that the partial pressure exerted by each gas is the product of its mole fraction and the total pressure, that is,
Pres. n-pentane = n x pressure(total)
1.92 = n x 1.99
n = 1.92/1.99 = 0.965 for n-pentane
For n-octane,
n = 1 - 0.965 = 0.035 for n-octane.
