<h3><u>Answer</u>;</h3>
1.0875 x 10-2 atm
<h3><u>Explanation;</u></h3>
2O3(g) → 3O2(g)
rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t
The average rate of disappearance of ozone ... is found to
be 7.25 × 10–3 atm over a certain interval of time.
This means (ignoring time)
∆[O3]/∆t = -7.25 × 10^–3 atm
(it is disappearing, thus the negative sign)
rate = -(1/2)∆[O3]/∆t
rate = -(1/2)*(-7.25 × 10^–3 atm)
= 3.625 × 10^–3 atm
Now use the other part of the expression:
rate = +(1/3)∆[O2)∆t
3.625 × 10–3 atm = +(1/3)∆[O2)/t
∆[O2)/∆t = (3)*(3.625× 10^–3 atm)
= 1.0875 x 10-2 atm over the same time interval
Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.
You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume
C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL
C1 x 250 mL = 0.2 M x 35 mL
C1 = (0.2 M x 35 mL) / 250 mL
C1 = 0.028 M of water added to 35mL of 0.2M HCl
Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
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Answer:
May be the instrument is incorrect or may be error in it.
Explanation:
The copper have not been detected by this test because the test may be not for the detection of copper, may be it is used for identification of another minerals. If there is copper in the lake sample but can't be detected in the test so it means that the instrument which is used for detection is not the right one or having error in that instrument. Every mineral has a specific type of instrument that detect its presence, if we use incorrect instrument for the mineral then we can't detect the presence of that specific mineral.
