This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]
hope this helps:)
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The answer is D only metals are shiny and highly malleable
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
<span>The half-life of 9 months is 0.75 years.
2.0 years is 2.0/0.75 = 2.67 half-lives.
Each half-life represents a reduction in the amount remaining by a factor of two, so:
A(t)/A(0) = 2^(-t/h)
where A(t) = amount at time t
h = half-life in some unit
t = elapsed time in the same unit
A(t)/A(0) = 2^(-2.67) = 0.157
15.7% of the original amount will remain after 2.0 years.
This is pretty easy one to solve. I was happy doing it.</span>
Answer:
A) potential energy is stored energy. Kenetic energy is energy of motion.
