isnt Q1 simply
VT=constant
solving we get Vf=47*295300
Please read the paragraph carefully and answer the following questions with proper explanations.
A student was investigating the speed of reaction between limestone granules and different concentrations of hydrochloric acid. However after doing a whole series of experiments at different acid concentrations, there was no time to do the last planned experiment. The volume of carbon dioxide collected after 5 minutes in a 100 cm3 gas syringe was used to determine the rate of reaction. All the experiments were done in one lesson at a temperature of 22 °C except for the last one. This was done in the next lesson, giving a carbon dioxide volume of 47.0 cm3 after 5 minutes, but at a higher temperature of 27 °C (when in Kelvin call this T1, and the other temperature T2).
To make the data analysis fair, all the gas volumes should be ideally measured at the same temperature, but a correction can be made for the last experiment.
1.The Volume the of 47.0 cm3 of gas at 27 °C, would occupy at 22 °C.
(a) 45.6 cm3
(b) 46.5 cm3
(c) 46.2 cm3
(d) 42.6 cm3
2.If the temperature was ignored, what is the % error in the rate of reaction measurement?
(a) +1.7%
(b) -1.7%
(c) +17%
(d) - 17%
3.Would you need to do any correction for the volume of acid added to the limestone?
(a) No correction needed for this at all.
(b) Yes correction need for this at all.
(c) Can’t be predicted .
(d) None of the above
For Q2 dont we need order of rate of reaction ???its not given...can u recheck plz
every question is a simple question ,if you understand well .
Q3
not sure..
for such a small temp diff...it weont make any significant diff on vol of acid...
so maybe a) [12]
how can that be possible ??
for A+B→C
rate of reaction =k[A]m[B]n
we need to knnow values of m and n to decide order of reaction and hence units..and hence error
or should we blindly take it anything...eg cm3/min
for q.3.
check it !!
Volume error = 47.0 - 46.2 = +0.8 cm3, therefore ....
% error = 0.8 x 100/47 = +1.7% (so you would over calculate the reaction rate
without this correction)
The % error in the volume would be the same as calculated for the rate e.g.
in cm3/min.