Answer:
<u>yes, this is an acceptable practice in research.</u>
<u>Explanation:</u>
Note that, having a different result in one's research does not imply it is automatically invalid. However, because there may have been some errors or limitations in the previous work by the male researcher, it is possible that the later results are more accurate, even though she used the same procedures. She may have been more careful when calculating her results.
Answer:
Chemical changes due to heat and pressure
Explanation:
Just took the test, hope this helps! :)
If we abbreviate the formula for nicotine as Nic, then the equations for two different equilibria of Nic in water are
Nic + H2O ---> NicH+ + OH-
NicH+ + H2O ---> NicH2 2+ + OH-
We can write the Kb1 expression for the first equation as
Kb1 = 1.0×10^-6 = [NicH+][OH-] / [Nic]
1.0×10^-6 = x^2 / 1.85×10^-3 - x
Approximating that x is negligible compared to 1.85×10^-3 simplifies the equation to
1.0×10^-6 = x^2 / 1.85×10^-3
x = 0.0000430
x = [OH-] = 4.30×10^-5 M
From the Kb2 expression
Kb2 = 1.3×10-11 = [NicH2 2+][OH-] / [NicH+]
1.1×10^-10 = x^2 / 4.30×10^-5 - x
Approximating that x is negligible compared to 4.30×10^-5 simplifies the equation to
1.1×10^-10 = x^2 / 4.30×10^-5
x = [OH-] = 6.88×10^-8
The concentration [OH-] can be computed as
[OH-] = 4.30×10^-5 M + 6.88×10^-8 M = 4.30×10^-5 M
This shows that the second equilibrium has a negligible effect on the pH.
We can now calculate for pH:
pOH = -log [OH-] = -log (4.30×10^-5 M) = 4.37
pH = 14 - pOH = 14 - 4.37 = 9.63
Answer:
29.8 mg
Explanation:
Just slightly less than 1/2 decay
