Answer:A theory is a well-substantiated explanation of an aspect of the natural world that can incorporate laws, hypotheses and facts.
Explanation:
Molarity is the number of moles of solute in one liter of solution whereas molality is the number of moles of solute present in 1 kilo gram of solution.
Molarity is denoted by M and molality denoted by m





Mass of solution

Density of solution = 1.1 g /mL
d = 1.1 = (W_B + W_A) /V

V = 910.11 mL
molarity of solution is:

= 0.0433 /0.910 = 0.0474 M
Thus, moalrity of solution is 0.0474 M
Answer:
It raises the boiling point and lowers the freezing point.
Explanation:
It is used as antifreeze in the cooling circuits of internal combustion engines, that is, it is used to reduce the melting point of the solution.
By adding ethylene glycol I'm not only bringing the melting point to -13°C, but the boiling point of ethylene glycol is 197°C.
since these substances not only lower the freezing point but also increase the boiling point, they are also called a colligative agent
<u>Solution</u> : Emperical Formula = 
Emperical Formula - A formula that gives the simplest whole number ratio of atom in a compound.
In the question percentages are given then assume that the total mass is 100grams so that the mass of each element is equal to the percentage given.
Given : 80.0g Carbon
20.0g Hydrogen
First convert the given masses into moles.


Value of each mole divided by the smallest number of moles and then round off.
Moles ratio of the element are


This is the mole ratio of the elements and is repersented by the subscript in the emperical formula.
Emperical Formula = 
Answer:
47 kJ·mol⁻¹
Explanation:
NaI(s) ⟶ Na⁺(aq) + I⁻(aq); ΔH = ?
We can carry out this process in a series of steps.
1. Convert the solid to its gaseous ions.
NaI(s) ⟶ Na⁺(g) + I⁻(g); -ΔH(lat) = 704 kJ
2. Hydrate the Na⁺ ions
Na⁺(g) ⟶ Na⁺(aq); ΔH(hyd) = - 410.0 kJ
3. Hydrate the I⁻ ions
I⁻(g) ⟶ I⁻(aq); ΔH(hyd) = -247
4. Add the equations
Cancel ions that occur on opposite sides of the reaction arrow
<u>ΔH/kJ·mol⁻¹</u>
NaI(s) ⟶ Na⁺(g) + I⁻(g); +704
Na⁺(g) ⟶ Na⁺(aq); - 410.0
I⁻(g) ⟶ I⁻(aq); -247
<u>NaI(s) ⟶ Na⁺(g) + I⁻(g)</u> <u> 47</u>
The heat of solution of NaI is 47 kJ·mol⁻¹.