The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.
Concentration is defined as the number of moles of a solute present in the specific volume of a solution.
According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.
M₁V₁=M₂V₂
Where, M₁=1.0M, V₁=14ml, M₂=?, V₂=25ml
Rearrange the formula for M₂
M₂=(M₁V₁/V₂)
Plug all the values in the formula
M₂=(1.0M×14 ml/25 ml)
M₂=14 M/25
M₂=0.56 M
Therefore, the concentration of a dextrose solution after the dilution is 0.56M.
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Answer:
704.6 g CO2
Explanation:
MM sucrose = 342.3 g/mol
MM CO2 = 44.01 g/mol
g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2
Answer:
An external stimuli is a stimulus that comes from outside an organism and causes a reaction.An internal stimuli is a stimulus that comes from inside an organism.
Testable, that is always required for a hypothesis. Otherwise, how can you prove it?
Answer:
2 mol NO2
Explanation:
3NO2(g)+H2O(l)→2HNO3(l)+NO(g)
from reaction 3 mol 1 mol
given 11 mol 3 mol
for 3 mol NO2 ----- 1 mol H2O
for x mol NO2 ----- 3 mol H2O
3:x = 1:3
x = 3 *3/1 = 9 mol NO2
So, for 3 mol H2O are needed only 9 mol NO2.
But we have 11 mol NO2. So, NO2 is in excess, and
11 mol NO2 - 9 mol NO2 = 2 mol NO2 will be left after reaction.
