Answer:
At end point there will a transition from pink to colorless.
Explanation:
As the student put the vinegar in the titrator and NaOH in the beaker, it means that he has poured phenolphthalein in the NaOH solution.
The pH range of phenolphthalein is 8.3-10 (approx), it means it will show pink color in basic medium.
So on addition of phenolphthalein in NaOH the solution will become pink in color.
When we start pouring vinegar from titrator neutralization of NaOH will begin.
On complete neutralization , on addition of single drop of vinegar the solution will become acidic and there will be complete disappearance of pink color solution in the beaker.
Answer:
In the third tube, the concentration is 0.16 ug/mL
Explanation:
In the first step, the solution is diluted by 5. Then, the concentration will be
20 ug/mL / 5 = 4 ug/mL
Then, in the second step this 4 ug / ml solution is diluted by a factor of five again:
4 ug /ml / 5 = 0.8 ug/mL
This solution is then diluted again by 5 and the concentration in the third tube will be then:
0.8 ug/mL / 5 = <u>0.16 ug/mL </u>
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Another way to calculate this is to divide the original concentration by the dilution factor ( 5 in this case) elevated to the number of dilutions. In this case:
Concentration in the third tube = 20 ug/mL / 5³ = 0.16 ug/mL
C, to make sure the design works as expected.
A prototype is first, typical model of the said product. Hope this helps!
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
Its either B or C. Its NOT A and D. I know that for a fact.
Hope this at least sorta helped. Have a good day! :D
