Answer:
Explanation:
To separate the a mixture of chalk and potassium chloride, we must not that chalk is calcium carbonate compound, CaCO₃.
Calcium carbonate has low solubility in water. KCl is readily soluble in water and it is also an ionic compound.
To separate a mixture of compounds with various solubility, we can carryout dissolution, filtration and evaporation.
We first pour pure water into the mixture. Water will dissolve the potassium chloride readily.
Then using a filter paper we filter out the suspended chalk particles. Leave the filtrate to then dry and collect it.
The solution filtered should be evaporated to dryness. This will leave the KCl behind from the solution.
Answer:
The answer to your question is 25 grams
Explanation:
Data
half-life = 5730 years
sample = 200 g
after 3 half-lives
Process
Calculate the amount of sample after one, two and three half-lives.
After each half-life, that of sample is half the previous amount.
Number of half-lives Amount of sample
0 200 g
1 100 g
2 50 g
3 25 g
Answer:
4 atmospheric pressure is needed to get a volume of 25 litres
Like dissolves like
so water is polar
CCl4 is nonpolar
LiCl is polar
CH4 is nonpolar
PCl6 is nonpolar
so LiCl would dissolve
The volume of base required to completely neutralize the acid is 3.2 mL of NaOH.
The equation of the reaction is;
2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)
From the question;
Concentration of acid CA = 0.426M
Concentration of base CB = 2.658M
Volume of acid VA = 10.00mL
Volume of base VB = ?
Number of moles of acid NA = 1
Number of moles of base NB = 2
Using the relation;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VB = CAVANB/CBNA
VB = 0.426M × 10.00mL × 2/ 2.658M × 1
VB = 3.2 mL
Learn more: brainly.com/question/6111443
