The molar mass (atomic weight ) of sodium is 23.0 grams/mole and the molar mass of sodium azide, NaN3 , is the mass of sodium, 23.0 gram/mole added to the molar mass of three atoms of nitrogen (14.0 x 3 = 42 gram/mole) which equals 65.0 grams/mole. The percentage of sodium is 23.0 /65.0 x 100 % = 35 %
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
A) The limiting reactant is Al
b) Br2 is the excess reactant
c) The amount moles of AlBr3 that get produced will be equal to the number of moles of Al to begin with.
d) 0
Answer:
- 602 mg of CO₂ and 94.8 mg of H₂O
Explanation:
The<em> yield</em> is measured by the amount of each product produced by the reaction.
The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.
The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

To calculate the yield follow these steps:
<u>1. Mole ratio</u>

<u />
<u>2. Convert 175mg of fluorene to number of moles</u>
- Number of moles = mass in grams / molar mass
<u>3. Set a proportion for each product of the reaction</u>
a) <u>For CO₂</u>
i) number of moles


ii) mass in grams
The molar mass of CO₂ is 44.01g/mol
- mass = number of moles × molar mass
- mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg
b) <u>For H₂O</u>
i) number of moles

ii) mass in grams
The molar mass of H₂O is 18.015g/mol
- mass = number of moles × molar mass
- mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
The molar mass of methylammonium bromide is 111u.
<h3>What is molar mass?</h3>
The molar mass is defined as the mass per unit amount of substance of a given chemical entity.
Multiply the atomic weight (from the periodic table) of each element by the number of atoms of that element present in the compound.
Add it all together and put units of grams/mole after the number.
Atomic weight of H is 1u
Atomic weight of N is 14u
Atomic weight of C is 12u
Atomic weight of Br is 79u
Calculating molar mass of
=2(1 x3+ 14+12+ 1 x 3 +79) = 111u
Hence, the molar mass of methylammonium bromide is 111u.
Learn more about molar mass here:
brainly.com/question/12127540
#SPJ1