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diamong [38]
3 years ago
10

Consider the reaction below. NH4+ + H2O mc013-1.jpg NH3 + H3O+ Which is an acid-conjugate base pair? NH4+ and NH3 NH4+ and H3O+

H2O and NH3 H2O and H3O+
Chemistry
2 answers:
juin [17]3 years ago
4 0
NH4+ and NH3 are an acid-conjugate base pair, since NH4+ is an acid, while NH3 is its conjugate base (since it is without the H+).
H2O and H3O+ can also be considered an acid-conjugate base pair, since H3O+ is an acid, while H2O would be its conjugate base. (But if only 1 answer is to be selected, it should be the NH4+ and NH3)
NH4+ and H3O+ are both acids, and both H2O and NH3 can be considered bases.
exis [7]3 years ago
3 0

Answer: A.NH4+ and NH3

Explanation:I took the test and got it correct

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3 years ago
calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?
inessss [21]

<u>Answer:</u>

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}

\Delta E=-1312[0.111-1]KJ mol^{-1}

\Delta E=1166 KJ mol^{-1}

\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}

=193623 \times 10^{-23}  \frac {J}{photon}

\Delta E=1.93623 \times 10^{-18}  \frac {J}{photon}

\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}

h is planck's constant  

c is the speed of light

λ is the wavelength of light  

\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18}  J/photon)}

Wavelength

\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m}  =103 nm (Answer)

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

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