Answer:
The mass was there all along, it was just in the air. The weight of the oxygen from the air is not weighed in the beginning, only at the end as part of the product, making it seem like there is a total mass change.
Answer:
15.70mg would remain
Explanation:
Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:
Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O
After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:
(19mg - X)
<em>Where X is the mass that now is in the aqueous phase</em>
Replacing in Kp formula:
9.5 = (19mg - X) / 5mL / (X /10mL)
0.95X = 19mg - X / 5mL
4.75X = 19 - X
5.75X = 19
X = 19 / 5.75
X = 3.30mg
That means 9-fluorenone that remain in the ether layer is:
19mg - 3.30mg =
<h3>15.70mg would remain</h3>
The sub-atomic particles of an atom are the proton, electron and the neutron. An electron has a charge of -1 and a smaller
mass than a proton. Proton has the same mass with the neutron. The ratio
between the mass of a proton and an electron is about 2000. An electron has an
equal value but negative charge with the proton.
<h3>
Answer:</h3>
= 5.79 × 10^19 molecules
<h3>
Explanation:</h3>
The molar mass of the compound is 312 g/mol
Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)
We are required to calculate the number of molecules present
We will use the following steps;
<h3>Step 1: Calculate the number of moles of the compound </h3>
Therefore;
Moles of the compound will be;
= 9.615 × 10⁻5 mole
<h3>Step 2: Calculate the number of molecules present </h3>
Using the Avogadro's constant, 6.022 × 10^23
1 mole of a compound contains 6.022 × 10^23 molecules
Therefore;
9.615 × 10⁻5 moles of the compound will have ;
= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules
= 5.79 × 10^19 molecules
Therefore the compound contains 5.79 × 10^19 molecules
Mass of water : 0.45 g
<h3>Further explanation</h3>
Given
1.5 g ethanoic acid
Required
Mass of water
Solution
Reaction
<em>CH₃COOH + NaOH ⇒H₂O + CH₃COONa</em>
mol CH₃COOH(MW=60 g/mol) :
= mass : MW
= 1.5 : 60
= 0.025
mol H₂O from the equation :
= 0.025(mol ratio CH₃COOH : H₂O= 1 : 1)
Mass of water :
= mol x MW water
= 0.025 x 18 g/mol
= 0.45 g
