<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.
<u>Explanation:</u>
We are given:
39.0 % of sodium in sodium chloride solution
This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution
Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)
Applying unitary method:
If 39 grams of sodium metal is present in 100 grams of sodium chloride solution
So, if 0.1 grams of sodium metal will be present in = 
of sodium chloride solution.
Hence, the mass of sodium chloride solution present is 0.256 grams.
Answer:
Esterification reaction
Explanation:
An esterification reaction is an organic reaction involving an organic acid and an alkanol to give an ester or an ethanoate and water
Like the name suggests, an ester is the product formed in an esterification reaction alongside water. It is like a neutralization reaction but this time it solely contains organic molecules. These molecules react with each other to give rise to another organic molecule which is a member of a different homologous series.
Practically, to form ethyl ethanoate, ethanoic acid react with ethanol in the presence of concentrated sulphuric acid which catalyses the reaction.
Answer:
<h3>The answer is 32 g/cm³</h3>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 768 g
volume = 24 cm³
We have
We have the final answer as
<h3>32 g/cm³</h3>
Hope this helps you
Answer:
2.3 * 10^-5
Explanation:
Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.
Hence;
Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols
From the question;
1.2 * 10^-3 mols dissolves in 250 mL
x moles dissolves in 1000mL
x = 1.2 * 10^-3 mols * 1000/250
x= 4.8 * 10^-3 moldm^-3
CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)
Hence Ksp = [Ca^2+] [C2O4^2-]
Where;
[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3
Ksp = (4.8 * 10^-3)^2
Ksp = 2.3 * 10^-5
Answer: well the receptacle connect the stalk to the flower and to support the flower and keeps the flower in an elevated position so as to attract the insects
Explanation: I don’t know if this helped
