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WITCHER [35]
3 years ago
13

A student conducting the iodine clock experiment accidentally makes an s2o32- stock solution that is too concentrated. how will

this affect the rate measurement? reaction 1: 3i- (aq) + s2o82- (aq) -> i3- (aq) + 2so42- (aq) slow reaction 2: i3- (aq) + 2s2o32- (aq) -> 3i- (aq) + s4o62- (aq) fast reaction 3: i3- (aq) + starch (aq) -> 3i- ---- starch (bluish black) fast
Chemistry
1 answer:
cupoosta [38]3 years ago
8 0
Adding (S2O3)2- would affect the reaction mechanism that involves this ion. From the reaction mechanism given above, the equilibrium of step 2 would be affected. Adding the stock solution of (S2O3)2- would shift the equilibrium to the right thus making more products of the said mechanism. Also, the reaction rate of this step would occur faster than the original rate. This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed. So, that after the addition, a color change would appear immediately because I3- would be removed slowly from solution, and would therefore be able to react with starch. 

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3 years ago
calculate the charge on plates if the plate area is 100 cm2 , the gap between plates is 10 mm and the scale is reading 1.0 gram.
Law Incorporation [45]

Answer: The charge on the plates are 88.4 picafarad

Explanation:The equation used in measuring charge in a plate is given as:

C=Q/V =E A/D

Where E= dielectric content

A= Area of plates

d= distance between plates

Using dielectric constant for Air=8.84×10-12F/m

A=100cm2=0.01m2

d=10mm=0.001m

C= 8.84×10-12×0.01/0.001

C= 88.4 picafarad

8 0
3 years ago
Determine whether a precipitate will form when 0.96g of Na2CO3 is combined with 0.20g BaBr2 in a 10 L solution.
Mashcka [7]

Answer:

Qsp > Ksp, BaCO3 will precipitate

Explanation:

The equation of the reaction is;

Na2CO3 + BaBr2 -------> 2NaBr + BaCO3

Since BaCO3 may form a precipitate we can determine the Qsp of the system.

Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles

concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M

Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles

concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M

Hence;

[Ba^2+] = 6.7 * 10^-5 M

[CO3^2-] = 9.1 * 10^-4 M

Qsp = [6.7 * 10^-5] [9.1 * 10^-4]

Qsp = 6.1 * 10^-8

But, Ksp for BaCO3 is 5.1*10^-9.

Since Qsp > Ksp, BaCO3 will precipitate

3 0
3 years ago
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