Nal is the sodium iodide
The oxidation number of sodium in sodium iodide is 1.<span>Formula: NaIHill system formula: I1Na1CAS registry number: [7681-82-5]Formula weight: 149.894Class: iodideColour: whiteAppearance: crystalline solidMelting point: 660°CBoiling point: 1304°CDensity: 3670 °C</span>
Answer:
i know it is alot but that is how are teacher told us to do, hope this is correct
Explanation:
a. 2.6
b. 12.0
Explanation:
a.
First, we will calculate the molar concentration of HCl.
M = mass of HCl / molar mass of HCl × liters of solution
M = 0.40 g / 36.46 g/mol × 4.5 L
M = 2.4 × 10⁻³ M
HCl is a strong monoprotic acid, so [H⁺] = 2.4 × 10⁻³ M. The pH is:
pH = -log [H⁺]
pH = -log 2.4 × 10⁻³ = 2.6
b.
First, we will calculate the molar concentration of NaOH.
M = mass of NaOH / molar mass of NaOH × liters of solution
M = 0.80 g / 40.00 g/mol × 2.0 L
M = 0.010 M
NaOH is a strong base with 1 OH⁻, so [OH⁻] = 0.010 M. The pOH is:
pOH = -log [OH⁻]
pOH = -log 0.010 = 2.0
The pH is:
14.00 = pH + pOH
pH = 14.00 - pOH = 14.00 - 2.0 = 12.0
Answer:
1) The vapor density of the organic compound is approximately 12.57
2) The relative molar mass (RMM) of the organic compound is approximately 25.14 grams
Explanation:
1) The mass of the balloon filled with dry hydrogen = 35 grams
The mass of the balloon filled with vapor of an organic compound = 440 grams
The vapor density = (Weight of a given volume of gas)/(Weight of equal volume of hydrogen)
The vapor density of the organic compound = (440)/(35) ≈ 12.57
The vapor density of the organic compound ≈ 12.57
2) The relative molar mass (RMM) = 2 × vapor density
The relative molar mass (RMM) of the organic compound = 2 × vapor density of the organic compound
The relative molar mass (RMM) of the organic compound ≈ 2 × 12.57 ≈ 25.14 grams
The relative molar mass (RMM) of the organic compound ≈ 25.14 grams