Match the following correctly. Boiled, deionized water Added to adjust the total

Question

Match the following correctly.

Boiled, deionized water

Added to adjust the total volume of each test soluton.

Buffer

Added to maintain a constant pH.

0.02M Na2S2O3

Added to show the appearance of I3- or depletion of thiosulfate ion.

Starch

Removes I3-as it is generated in the reaction.

Rate law

k[A2]p[B2]q

p

Determined from slope of plot of log (rate) vs log [I-]

q

Determined from slope of plot of log (rate) vs log [H2O2]

Given that 4.0x10-5 moles of S2O32- were consumed, calculate the Rate for the kinetic trial knowing that it took 89 seconds to react.

2S2O32-(aq) + I3-(aq) 3 I -(aq) + S4O62-(aq)

Tip: Use your blackboard notes/manual if needed.

(Provide your answer in scientific notation and with two sig figs) Example: if answer is 0.00004 then input 4.0e-5 or 4.0E-5

Given the following data, calculate the (mol I3-) produced.

Moles of S2O32-, consumed: 0.00014

2S2O32-(aq) + I3-(aq) 3 I -(aq) + S4O62-(aq)

Hint: Use a mole to mole ratio from the balanced equation.

Tip: Use your blackboard notes/manual if needed.

(Provide your answer in scientific notation and with two sig figs.) Example: if answer is 0.000065 then input 6.5e-5 or 6.5E-5

Match the following correctly.

Question Selected Match

Boiled, deionized water

B.

Added to adjust the total volume of each test soluton.

Buffer

G.

Added to maintain a constant pH.

0.02M Na2S2O3

E.

Added to show the appearance of I3- or depletion of thiosulfate ion.

Starch

C.

Removes I3-as it is generated in the reaction.

Rate law

A.

k[A2]p[B2]q

p

F.

Determined from slope of plot of log (rate) vs log [I-]

q

D.

Determined from slope of plot of log (rate) vs log [H2O2]

Explanation / Answer

Given that 4.0x10-5 moles of S2O32- were consumed, calculate the Rate for the kinetic trial knowing that it took 89 seconds to react.

dC / dT = (4*10^-5)/(89) = 4.49438*10^-7 M/s

Given the following data, calculate the (mol I3-) produced. Moles of S2O32-, consumed: 0.00014

2S2O32-(aq) + I3-(aq) 3 I -(aq) + S4O62-(aq)

for 1 mol of Se23-2 we react 1/2 mol of I3-

so

for 0.00014 moles consumed, then 0.00014/2 =0.00007 mol of I2 are consumed = 7*10^-5

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