Answer is: D. Na2SO4.
b(solution) = 0.500 mol ÷ 2.0 L.
b(solution) = 0.250 mol/L.
b(solution) = 0.250 m; molality of the solutions.
ΔT = Kf · b(solution) · i.
Kf - the freezing point depression constant.
i - Van 't Hoff factor.
Dissociation of sodium sulfate in water: Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq).
Sodium sulfate dissociates on sodium cations and sulfate anion, sodium sulfate has approximately i = 3.
Sodium chloride (NaCl) and potassium iodide (KI) have Van 't Hoff factor approximately i = 2.
Carbon dioxide (CO₂) has covalent bonds (i = 1, do not dissociate on ions).
Because molality and the freezing point depression constant are constant, greatest freezing point lowering is solution with highest Van 't Hoff factor.
If it’s hydraulic turbine then it’s potential and kinetic energy and if it’s a thermal process then heat energy from the fuel burnt runs the turbine
Ionic compounds are formed between oppositely charged ions.
A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).
To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.
First empirical formula of binary ionic compound is written between
First Formula would be 
Second empirical formula is between 
Second Formula would be 
Note : When the subscript are same they get cancel out, so
would be written as 
Third empirical formula is between 
Third Formula would be :
Forth empirical formula is between 
Forth Formula would be :
or 
Note- The subscript will be simplified and the formula will be written as
.
The empirical formula of four binary ionic compounds are : 
Answer:
3
Explanation:
Applying,
= R/R'............... Equation 1
Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.
From the question,
Given: R = 40 atoms, R' = 5 atoms
Substitute these values into equation 1
= 40/5
= 8
= 2³
Equation the base,
n' = 3