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sertanlavr [38]
3 years ago
14

A 100. ml portion of 0.250 m calcium nitrate solution is mixed with 400. ml of 0.100 m nitric acid solution. what is the final c

oncentration of the nitrate ion?
Chemistry
1 answer:
pshichka [43]3 years ago
6 0
<span>Answer: Moles Ca(NO3)2 = 100 x 0.250 / 1000 = 0.025 Ca(NO3)2 >> Ca2+ + 2NO3- Moles NO3- = 2 x 0.025 = 0.05 Moles HNO3 = 400 x 0.100 / 1000 = 0.04 Total moles = 0.05 + 0.04 = 0.09 Total volume = 500 ml = 0.500 L M = 0.09 / 0.500 = 0.18</span>
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Answer :  The energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

Explanation : Given,

Wavelength = 435.8nm=435.8\times 10^{-9}m

conversion used : 1nm=10^{-9}m

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency

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Now put all the given values in the above formula, we get:

E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}

E=4.56\times 10^{-19}J

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

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