which gas law would you use to calculate the new pressure for a gas when the volume of the trapped gas changes while the tempera
2 answers:
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Answer:
His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.
Explanation:
The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.
Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.
Suppose the mass of the solute is m.
Originally, the density is =
Now after adding extra 10 mL , the density becomes .
Therefore,
So the density decreases when we add more solution.
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
=14-56.17
=8.823
The volume of the = 40.00 ml
convert into the liter= 0.040L
The value of the concentrated =0.10 M
The volume of the = 50.00 ml
convert into the liter= 0.050L
The value of concentrated = 0.10 M
The volume of the = 30 ml
convert into the liter= 0.030L
The value of concentrated =0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of =
calculating the new concentrated value of =
calculating the new concentrated value of
when 1 mol
produced 2 mols
so, 0.0125 in
produced:
create the ICE table:
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83: