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: 48800g
Explanation:
Using the mathematical relation : Moles = Mass / Molar Mass
Moles = 488
Molar mass of CaCO3 = 40 + 12 + (16 x 3) = 100g/mol
Therefore
488 = mass / 100 = 48800g
The patient needs 1000 ml of 5% (w/v) glucose solution
i.e 1000 ml x 5 g/ 100 ml
where the stock solution is 55% (w/v) = 55 g / 100 ml
So, 1000 ml x 5 g / 100 ml = V (ml) x 55 g / 100 ml
V = 1000 x (5 / 100) / (55 / 100) = 5000 / 55 = 90.9 ml
∴ the patient needs 90.9 ml of 55% (w/v) glucose solution
I do not know your lines are all messed up is that 2 words after matter or just 1
The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 
10^-5 V.
<u>Explanation:</u>
Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.
Given p = 1.47 D = 1.47 3.34 10^-30 = 4.90 10^-30.
V = 1 / (4π∈о) (p cos(θ)) / (r^2)
where p is a permanent electric dipole,
∈ο is permittivity,
r is the radius from the axis of a dipole,
V is the electric potential.
V = 1 / (4 3.14 8.85 10^-12) (4.90 10^-30 1) / (55.3 10^-9)^2
V = 1.44 10^-5 V.
