The concentration of [H3O⁺]=2.86 x 10⁻⁶ M
<h3>Further explanation</h3>
In general, the weak acid ionization reaction
HA (aq) ---> H⁺ (aq) + A⁻ (aq)
Ka's value
![\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKa%20%5C%3A%20%3D%20%5C%3A%20%5Cfrac%20%7B%5BH%20%5E%20%2B%5D%20%5BA%20%5E%20-%5D%7D%20%7B%5BHA%5D%7D%7D%7D%7D)
Reaction
HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵
![\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}](https://tex.z-dn.net/?f=%5Ctt%20Ka%3D%5Cdfrac%7B%5BC_2H_3O%5E%7B2-%7D%5BH_3O%5E%2B%5D%5D%7D%7B%5BHC_2H_3O_2%5D%7D%7D%5C%5C%5C%5C1.8%5Ctimes%2010%5E%7B-5%7D%3D%5Cdfrac%7B0.22%5Ctimes%20%5BH_3O%5E%2B%5D%7D%7B0.035%7D)
[H₃O⁺]=2.86 x 10⁻⁶ M
Answer: -
²²³Ra is the daughter nuclide produced when ²²⁷Th undergoes alpha decay
Explanation: -
When alpha decay occurs, the mass number of the parent decreases by 4 and the atomic number decreases by 2.
Mass number of ²²⁷Th =227
Atomic number of ²²⁷Th = 90
Mass number of daughter = 227 - 4 = 223
Atomic number of daughter = 90 - 2 = 88
88 is the atomic number of Ra Radium.
Thus the daughter is ²²³Ra
Answer:
The correct answer is:
An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate.
Explanation:
In fact, violet has higher frequency than green light. This means that photons on violet carry more energy than photons of green light (remember that the energy of a photon is proportional to it's frequency:

, so when they hit the surface of the metal, more energy is transferred to the electrons. The electron was already emitted with green light, so it must be emitted with also violet light, given the more energy transferred.