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e-lub [12.9K]
3 years ago
5

The carbon dioxide exhaled by astronauts can be removed from a spacecraft by reacting it with lithium hydroxide (LiOH). The reac

tion is as follows: CO2 (g) + 2LiOH (s) Li2CO3 (s) + H2O (l). An average person exhales about 20 moles of CO2 per day. How many moles of LiOH would be required to maintain two astronauts in a spacecraft for three days?
Chemistry
1 answer:
maks197457 [2]3 years ago
8 0
Two astronauts would exhale about 40 moles of carbon dioxide daily.

Carbon dioxide reacts with lithium hydroxide in a 1 : 2 mole ratio. Set up a proportion:

 1 : 2 = 40 : x

Then, find x: <span>12=40x

</span> Cross multiply. x = 80 moles of LiOH per day for both astronauts
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What are the four categories commonly used to classify chemicals and chemical agents in the graphic communications industry
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Answer:

Organic

Plate making

Ink mists

Gas, fumes and dust

Explanation:

the four categories commonly used to classify chemicals and chemical agents in the graphic communications industry are Organic ,Plate making ,Ink mists, Gas, fumes and dust.

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5 0
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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3 years ago
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