Answer:
a) 38.2 % mass
b) 61.8 g solute/100 g solvent
c) 1.65 g/mL
Explanation:
Given the data:
mass of solute = 17.5 g
mass of solvent= 28.3 g
total solution volume= 27.8 mL
a)- mass percent= mass of solute/mass of solution x 100
mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g
mass % = 17.5 g/45.8 g x 100 = 38.2 % mass
b)- solubility = grams of solute/ 100 g solvent
= 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent
c)- density = massof solution/total volumesolution = 45.8 g/27.8 mL = 1.65 g/mL
Answer:
Conduct electricity when they are molten, while covalent compounds usually do not conduct electricity when they are molten.
Entropy Change is calculated by (Energy transferred) / (Temperature in kelvin)
deltaS = Q / T
Q = (mass)(latent heat of fusion)
Q = m(hfusion)
Q = (500g)(333J/g) = 166,500J
T(K) = 32 + 273.15 = 305.15K
deltaS = 166,500J / 305.15K
deltaS = 545.63 J/K
Answer : q = 6020 J, w = -6020 J, Δe = 0
Solution : Given,
Molar heat of fusion of ice = 6020 J/mole
Number of moles = 1 mole
Pressure = 1 atm
Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.
The relation between heat and molar heat of fusion is,
(in terms of mass)
or, 
(in terms of moles)
Now we have to calculate the value of q.
When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.
So, the value of 
Now we have to calculate the value of w.
Formula used : 
where, q is heat required, w is work done and 
is internal energy.
Now put all the given values in above formula, we get
w = -6020 J
Therefore, q = 6020 J, w = -6020 J, Δe = 0
To find average atomic mass you multiply the mass of each isotope by its percentage, and then add the values up.
35 * 0.90 + 37 * 0.08 + 38 * 0.02 = 35.22
Average atomic mass closest to 35.22 amu.
