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(Solved): I'm stuck on this last one. Took a bigger screenshot in case other parts are relevant. d) If t ...



I'm stuck on this last one. Took a bigger screenshot in case other parts are relevant.

d) If the initial volume of the H2O2H2O2 solution is 1.6 LL , what total amount of O2O2 (in moles) is formed in the first 50 ss of reaction?  

\( - \) Part C
Calculate the instantaneous rate of formation of \( \mathrm{O}_{2} \) at \( 50 \mathrm{~s} \).
Express your an

\( - \) Part C Calculate the instantaneous rate of formation of \( \mathrm{O}_{2} \) at \( 50 \mathrm{~s} \). Express your answer using two significant figures. \( \checkmark \) Correct The instantaneous rate of a reaction is the rate at any one point in time. You can determine the instantaneous rate by calculating the slope of the line tangent to the curve at the point of interest. The instantaneous rate in terms of \( \left[\mathrm{O}_{2}\right] \) is instantaneous rate \( ( \) at \( 50 \mathrm{~s})=\frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t}=-\frac{1}{2} \frac{\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}=-\frac{1}{2} \frac{0.15 \mathrm{M}-0.30 \mathrm{M}}{60 \mathrm{~s}-40 \mathrm{~s}}=3.8 \times 10^{-3} \mathrm{M} / \mathrm{s} \) - Part D If the initial volume of the \( \mathrm{H}_{2} \mathrm{O}_{2} \) solution is \( 1.6 \mathrm{~L} \), what total amount of \( \mathrm{O}_{2} \) (in moles) is formed in the first \( 50 \mathrm{~s} \) of reaction? Express your answer using two significant figures. Figure \( \times \) Incorrect; Try Again; \( \mathbf{5} \) attempts remaining


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