The molecule is thiophosgene or CSCl₂.
<h3><u>Explanation:</u></h3>
As the informations given, the molecule contains 12.17 grams of carbon, 32.53 g of sulfur, and 72.19 g of chlorine.
Molar mass of carbon =12.
Molar mass of sulphur =32.
Molar mass of chlorine =35.
So on an average estimation, one mole of the molecule contains one mole of carbon atoms, one mole of Sulphur atoms and two moles of chlorine atoms. Molar mass of an element is the average weight of one mole of atoms of that element which is made average according to the variations of percentage availability of isotopes of that element on earth.
So the estimated molecule is CSCl₂ or thiophosgene.
Answer:
Explanation:
Hello,
By considering the rate law:
![ln(\frac{[CO_2]}{[CO_2]_0} )=-kt\\](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BCO_2%5D%7D%7B%5BCO_2%5D_0%7D%20%29%3D-kt%5C%5C)
Solving for the time, we've got:
![k=-\frac{ln (\frac{[CO_2]}{[CO_2]_0 })}{t}\\\\k=-\frac{ln(\frac{56.0mmol*dm^{-3} }{220mmol*dm^{-3} })}{1.22x10^{4}s }\\ \\k=1.12x10^{-4}s](https://tex.z-dn.net/?f=k%3D-%5Cfrac%7Bln%20%28%5Cfrac%7B%5BCO_2%5D%7D%7B%5BCO_2%5D_0%20%7D%29%7D%7Bt%7D%5C%5C%5C%5Ck%3D-%5Cfrac%7Bln%28%5Cfrac%7B56.0mmol%2Adm%5E%7B-3%7D%20%7D%7B220mmol%2Adm%5E%7B-3%7D%20%7D%29%7D%7B1.22x10%5E%7B4%7Ds%20%7D%5C%5C%20%20%5C%5Ck%3D1.12x10%5E%7B-4%7Ds)
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<span>A population of the frogs that do not receive the UV radiation would work as a good control group. In this case, since the impact of the radiation is the variable under study, a group that receives no radiation, and thereby should not exhibit a change, would be a control group that could be compared against the frogs that do receive the radiation.</span>
Explanation:
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