Answer: 
Explanation:
Depression in freezing point is:
where,
= freezing point of solution = ?
= freezing point of solvent (cyclohexane) = 
= freezing point constant of solvent (cyclohexane) = 
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
= mass of solute (biphenyl) = 0.771 g
= mass of solvent (cyclohexane) = 25.0 g
= molar mass of solute (biphenyl) =
Now put all the given values in the above formula, we get:
Therefore, the freezing point of a solution made by dissolving 0.771 g of biphenyl in 25.0 g of cyclohexane is
Cu (s) + 2AgNO3 ---> 2Ag + Cu(NO3)2
=13.5 moles Silver Nitrate
<u>Answer:</u> The empirical and molecular formula of the compound is 
and 
respectively
<u>Explanation:</u>
We are given:
Mass of C = 3.758 g
Mass of H = 0.316 g
Mass of O = 1.251 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.
For Carbon = 
For Hydrogen = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 4 : 4 : 1
The empirical formula for the given compound is
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
We are given:
Mass of molecular formula = 130 g/mol
Mass of empirical formula = 68 g/mol
Putting values in above equation, we get:
Multiplying this valency by the subscript of every element of empirical formula, we get:
Hence, the empirical and molecular formula of the compound is and respectively
Answer:
It is called Potassium Hydroxide
Answer:
Döbereiner singled out triads of the elements lithium, sodium and potassium as well as chlorine, bromine and iodine. He noticed that if the three members of a triad were ordered according to their atomic weights, the properties of the middle element fell in between those of the first and third elements.
Explanation:
