Eutrophication results from an excess of
a. phosphates
The question is incomplete, here is the complete question:
Calculate the mole fraction of the ionic species KCl in the solution A solution was prepared by dissolving 43.0 g of KCl in 225 g of water.
<u>Answer:</u> The mole fraction of KCl in the solution is 0.044
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of water = 225 g
Molar mass of water = 18 g/mol
Putting values in equation 1, we get:

Given mass of KCl = 43 g
Molar mass of KCl = 74.55 g/mol
Putting values in equation 1, we get:

Mole fraction of a substance is given by:

Moles of KCl = 0.577 moles
Total moles = [0.577 + 12.5] = 13.077 moles
Putting values in above equation, we get:

Hence, the mole fraction of KCl in the solution is 0.044
Answer:
3.657 kg
Explanation:
Given:
Enthalpy of combustion of hard coal = -35 kJ/g
Enthalpy of combustion of gasoline = 1.28 × 10⁵ kJ/gal
Density of gasoline = 0.692 g/mL
now,
The heat provide 1 gallon of gasoline provide = 1.28 × 10⁵ kJ
and,
heat provided by the 1 gram of coal = 35 kJ
or
1 kJ of heat is provided by (1/35) gram of hard coal
therefore,
For 1.28 × 10⁵ kJ of heat, mass of hard coal = 1.28 × 10⁵ kJ × (1 / 35)
or
For 1.28 × 10⁵ kJ of heat, mass of hard coal = 3657.14 grams = 3.657 kg
Answer:
Mass number -Atomic number