Extra Credit 33
- Page ID
- 83266
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
Q17.5.1
An electric battery is safe to use and has a constant voltage with high current output.
Q12.1.6
Rate=-Δ[Br-]/5Δt
=-Δ[BrO3-]/Δt
=-Δ[H+]/6Δt
=Δ[Br2]/3Δt
rate=-[Br-]/5Δt
=Δ[Br2]/3Δt
so, Δ[Br2]/ Δt=2.1*10^(-4) Ms^(-1)
Q12.5.5
-Ea/R=ln(k) vs. 1/T
Q21.3.8
Moles of N2=100gC*(1 molC/12gC)*(1 molN2/1 molC)=8.33 molN2
PV=nRT
(1atm)(V)=(8.33 molN2)(0.008206 L*atm/(mol*k))(273K)
V=18.66 L of N2 gas
Q20.2.4
Oxidized is when an atom loses electrons. Reduced is when an atom gains electrons.
Example: CH4+O2➡H2O+CO2
Q20.4.23
E°cell=E°cathode-E°anode=0.54-（-0.76）=1.30V
Q20.9.8
Cu2+ +2e- ➡Cu
5.12A*1.5 h*(3600s/1h)*(1 mol e-/96485C)*(1 mol Cu(NO3)2/2 mol e-)*(1 mol Cu/1 mol Cu(NO3)2)*(63.546 g Cu/1 mol Cu)=9.105 g of Cu
Q14.6.6
The second step determines the rate constant of the overall reaction because this is the slowest step (requires a lot of energy to react).