Common salt
sulfur dioxide
Answer: 
Explanation:
where,
= boiling point of solution = ?
= boiling point of solvent (X) = 
= freezing point constant = 
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte like urea)
= mass of solute (urea) = 29.82 g
= mass of solvent (X) = 500.0 g
= molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
Therefore, the freezing point of solution is 
Answer:
density of a piece of metal = 7 gr/ml
Explanation:
See the file please
Answer:
a. C: +3 ; b. N: +5 ; c. S:+6 ; d. C: +4; e. Mn: +7 ; f. Cr: +6.
Explanation:
Global charges in molecules is 0
You sum all the oxidation states to determine the oxidation state for the compound.
Na₂C₂O₄ → Sodium oxalate → Global charge: 0
Oxidation state for C: +3
HNO₃ → Nitric acid → Global charge: 0
Oxidation state for N: +5
H₂SO₄ → Sulfuric acid → Global charge: 0
Oxidation state for S: +6
HCO₃⁻ → Bicarbonate → Global charge: -1, this is an anion
Oxidation state for C: +4
KMnO₄ → Potassium permanganate → Global charge: 0
Oxidation state for Mn: +7
Cr₂O₇⁻ → Anion dichromate → Global charge: -2
Oxidation state for Cr: +6
Vegetative propagation is a form of asexual reproduction of a plant. Only one plant is involved and the offspring is the result of one parent. The new plant is genetically identical to the parent. i hope this helped <3
