Rate = k [H+ja [1-16 [starch-] [H2O2] - N.b. It is expected that at least one of these exponents will be 0. Test for 1

[answerhappygod]

Rate K H Ja 1 16 Starch H2o2 N B It Is Expected That At Least One Of These Exponents Will Be 0 Test For 1 1 (3.76 KiB) Viewed 66 times

Rate K H Ja 1 16 Starch H2o2 N B It Is Expected That At Least One Of These Exponents Will Be 0 Test For 1 2 (30.28 KiB) Viewed 66 times

Rate K H Ja 1 16 Starch H2o2 N B It Is Expected That At Least One Of These Exponents Will Be 0 Test For 1 3 (31.84 KiB) Viewed 66 times

Rate K H Ja 1 16 Starch H2o2 N B It Is Expected That At Least One Of These Exponents Will Be 0 Test For 1 4 (31.9 KiB) Viewed 66 times

Rate K H Ja 1 16 Starch H2o2 N B It Is Expected That At Least One Of These Exponents Will Be 0 Test For 1 5 (30.75 KiB) Viewed 66 times

Rate K H Ja 1 16 Starch H2o2 N B It Is Expected That At Least One Of These Exponents Will Be 0 Test For 1 6 (3.76 KiB) Viewed 66 times

I need help calculating the rate law for each using the averages
from trials 3 and 4 and 5and6. I havre calculated everything. I
just need help with calculating the rate law
Rate = k [H+ja [1-16 [starch-] [H2O2] - N.b. It is expected that at least one of these exponents will be 0.
Test for 1- Trials I- H2O H+ S2032- H202 Starch Total drops of solution Time in second s Rate (1/time) Average Rate Trial 1 4 5 1 4 4 2 20 99 0.010 Trial 2 4 5 1 4 4 2 20 98 0.010 0.010 Trial 3 2 7 1 4 4 2 20 105 0.009 Trial 4 2 7 1 4 2 20 176 0.006 0.0075 | Trial 5 8 1 1 4 4 2 20 52 0.019 Trial 6 8 1 1 4 4 2 20 68 0.015 0.017
Test for H+ Trials I- H20 H+ S2032- H202 Starch Total drops of solution Time in second s Rate (1/time) Average Rate Trial 1 4 5 1 4 4 2 20 93 0.011 Trial 2 4 5 1 4 4 2 20 96 0.010 0.010 Trial 3 4 4 2 4 4 2 20 61 0.016 Trial 4 4 4 2 4 2 20 60 0.017 0.0165 Trial 5 4 2 4 4 4 2 20 78 0.013 Trial 6 4 2 4 4 4 2 20 54 0.019 0.016
Test for Starch Trials I- H2O H+ S2032- H202 Starch Total drops of solution Time in second s Rate (1/time) Average Rate Trial 1 4 5 1 4 2 20 102 0.010 Trial 2 4 5 1 4 4 2 20 92 0.011 0.010 Trial 3 4 3 1 4 4 4. 20 100 0.01 Trial 4 4 3 1 4 4 20 109 0.009 0.0095 Trial 5 4 0 1 4 4 7 20 110 0.009 Trial 6 4 0 1 4 4 7 20 150 0.007 0.008
Test for H202 Trials I- H2O H+ S2032- H202 Starch Time in second Total drops of solution Rate (1/time) Average Rate S Trial 1 4 5 1 4 4 2 20 76 0.013 Trial 2 4 5 1 2 20 91 0.011 0.012 Trial 3 4 7 1 4 2 2 20 107 0.009 Trial 4 4 7 1 4 2. 2 20 122 0.008 0.0085 Trial 5 4 8 8 1 4 1 N 20 460 0.002 Trial 6 4 8 1 4 1 2 20 481 0.002 0.002
Rate = k [H+ja [1-16 [starch-] [H2O2] - N.b. It is expected that at least one of these exponents will be 0.