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:
Its the bottom one :) ......
Answer:
A. an equal and opposite reaction force
Explanation:
This means that there is a natural reaction of force
Answer:
It's false ok it's non electrolyte
<h3>
Answer:</h3>
1.2 × 10^-11 M
<h3>
Explanation:</h3>
<u>We are given;</u>
Concentration of Hydrogen ions [H⁺] as 0.00083 M
<u>We are required to calculate the concentration of [OH⁻]</u>
We know that;
pH = - log [H⁺]
POH = -log[OH⁻]
Also, pH + pOH = 14
With the concentration of H⁺ we can calculate the pH
pH = -log 0.00083 M
= 3.08
But; pH + pOH = 14
Therefore; pOH = 14 - pH
= 14 - 3.08
= 10.92
But, pOH = -log[OH⁻]
Therefore; [OH] = -Antilog(pOH)
Hence; [OH⁻] = Antilog -10.92
= 1.2 × 10^-11 M
Therefore, [OH⁻] is 1.2 × 10^-11 M
