Divide the percentages by the RAM and then divide by the smallest value, this will give a ratio of the elements in the compound. N =14, O=16
Nitrogen : Oxygen
36.84/14 : 63.16/16
2.5 : 3.9
Divide both values by the lowest, and try to find whole number ratios:
2.5/2.5 : 3.9/2.5
1 : 1.5
Since the ratio of oxygen to nitrogen is 1.5 : 1, we have to double it so that it become 3:2
The empirical formula would be N₂O₃
a) First, to get ΔG°rxn we have to use this formula when:
ΔG° = - RT ㏑ K
when ΔG° is Gibbs free energy
and R is the constant = 8.314 J/mol K
and T is the temperature in Kelvin = 25 °C+ 273 = 298 K
and when K = 4.4 x 10^-2
so, by substitution:
ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)
= -7739 J = -7.7 KJ
b) then, to get E° cell for a redox reaction we have to use this formula:
ΔE° Cell = (RT / nF) ㏑K
when R is a constant = 8.314 J/molK
and T is the temperature in Kelvin = 25°C + 273 = 298 K
and n = no.of moles of e- from the balanced redox reaction= 3
and F is Faraday constant = 96485 C/mol
and K = 4.4 x 10^-2
so, by substitution:
∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)
= - 2.7 x 10^-2 V
The answer is Selenium.
Hope that helps.
Answer:
nitrogen is the answer because of the same position of group in periodic table
<span>Answer:
.01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L.
so .01/.015 to .005/.015 ~ 67% D to 33% L.
And thus, the enantiomer excess will be 34%.</span>
