And how do we know? Well, by experiment, and measurement, but VESPER gives a good approximation. We have $2xx5+6=16$ $"valence electrons"$ to distribute over 3 centres. A structure of $N-=N^(+)-O^(-)$...
1 Answers 1 viewsIn finding the amount of moles there are, you first have to know what is and what it represents. Avogadro's number is $6.022$x$10^23$. That number can represent anything; in chemistry...
1 Answers 1 viewsNote: This answer may seem to be quite long, but that is because I have tried to explain each of the steps fully and clearly. If you want, you can...
1 Answers 1 views$"Moles of nitrous oxide"=(85*g)/(44.01*g*mol^-1)$ $=1.93*mol$. Since, clearly, there are $"2 moles of nitrogen atoms"$ $"per mole of nitrous oxide"$, we simply double the molar quantity of $"nitrous oxide:"$ $1.93*molxx2*mol*N*mol^-1NO=??mol$
1 Answers 1 viewsFrom (PV = nRT) $P = (nRT/V) = (R("mass"/"f.wt")(T/V))$ $P = R("mass"/"Volume")(T/"f.wt.")$= $R("Density")(T/"f.wt.")$ Given: $R = 0.08206("L-Atm"/"mol-K")$ $"Density" = 2.85(g/L)$ $T = 298K$ $f.wt. = 44(g/"mol")$ $P = (0.08206"L-Atm"/"mol-K")(2.85g/L)(298K)(44g/"mol")^-1 =...
1 Answers 1 views$"Dalton's Law of Partial Pressures"$ states UNEQUIVOCALLY that in a gaseous mixture, the partial pressure exerted by a gaseous component is the SAME as it would exert if it ALONE...
1 Answers 1 views$"Dalton's Law of partial pressures"$ holds that in a gaseous mixture, the total pressure is the sum of the individual partial pressures; i.e. the sum of the pressures of the...
1 Answers 1 viewsI got $"71.4 s"$, after we have gone through the two half-lives for $"NO"_2$. This seems to be a second-order half-life (yes, that is a thing). I don't know...
1 Answers 1 viewsYou used the so-called $"Mohr's salt"$, which is a stable and convenient source of $"Fe(II) salt"$, i.e. $[(NH_4)_2Fe(SO_4)_2]*6H_2O$. We can ignore the ammonium and sulfate shrubbery, and represent the redox...
1 Answers 1 views$"Oxidation half equation:"$ $NO_2(g) +H_2O(l) rarr NO"_3^(-) +2H^+ + e^(-) $ $stackrel(+IV)Nrarrstackrel(+V)N$ Charge and mass are balanced so tick. $"Reduction half equation:"$ $Cr_2O_7^(2-)(aq) +14H^(+) +6e^(-) rarr 2Cr^(3+) +7H_2O $ $stackrel(VI+)"Cr"rarrstackrel(+III)"Cr"$...
1 Answers 1 views