The hydrogen one because who doesnt like snoballs'
1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.
2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.
3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.
4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.
5. If the hypothesis turns out to be true, then all the mixtures prepared should freeze and become solid after a certain period of time, with the exception of the 50:50 mixture. The 50:50 mixture should still remain as a liquid even when left overnight.
Answer:
2.57 e-9
Explanation:
The formula is H3O=10^-Ph
10^-8.59=2.57 e-9
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.
To know more about dilution
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Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
