Answer:
empirical formula = C3H7
molecular formula = C6H14
You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!
group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1
we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!
Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.
A solvent (from the Latin solvō, "I loosen, untie, I solve") is a substance that dissolves a solute (a chemically distinct liquid, solid or gas), resulting in a solution. A solvent is usually a liquid but can also be a solid or a gas. The quantity of solute that can dissolve in a specific volume of solvent varies with temperature. Common uses for organic solvents are in dry cleaning (e.g., tetrachloroethylene), as paint thinners (e.g., toluene, turpentine), as nail polish removers and glue solvents (acetone, methyl acetate, ethyl acetate), in spot removers (e.g., hexane, petrol ether), in detergents (citrus terpenes) and in perfumes (ethanol). Water is a solvent for polar molecules and the most common solvent used by living things; all the ions and proteins in a cell are dissolved in water within a cell. Solvents find various applications in chemical, pharmaceutical, oil, and gas industries, including in chemical syntheses and purification processes.
source: wikapedia
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
Answer:
Mass of solution=100g
mass of salt=20g
so; mass of solute=80g
percentage composition =(mass of salt/total
mass) ×100
= \frac{20}{100} \times 100 \\ = 20\%
glad to help you
hope it helps
