Asked by Amanda
Use the concentrations and volumes of hydrogen peroxide and KI before mixing (Table 1),
calculate the post-mixing concentrations.
Before mixing:
[H2O2] = 0.88M
[KI] = 0.500M
Table 1: before mixing
1. Vol H2O2 = 4.0 mL, [H2O2] = 0.88M
2. Vol KI = 1.0 mL, [KI] = 0.500 M
units: M
I have kept trying to do this one but I am not getting the right answer. the answer is given (after mixing), [H2O2]=0.70M, [KI]=0.100M
calculate the post-mixing concentrations.
Before mixing:
[H2O2] = 0.88M
[KI] = 0.500M
Table 1: before mixing
1. Vol H2O2 = 4.0 mL, [H2O2] = 0.88M
2. Vol KI = 1.0 mL, [KI] = 0.500 M
units: M
I have kept trying to do this one but I am not getting the right answer. the answer is given (after mixing), [H2O2]=0.70M, [KI]=0.100M
Answers
Answered by
chioma
i donot know this one next
Answered by
DrBob222
[H2O2] before = 0.88M
volume H2O2 before mix is 4 mL
total volume after mix is 5 (4 from H2O2 + 1 from KI) so the H2O2 has been diluted from 4 to 5 mL.'
0.88 x 4/5 = ?
KI before mix is 0.500.
volume KI before mix is 1 mL.
total volume after mix is 5 (1 from KI + 4 from H2O2). The 0.500 M KI has been diluted from 1 mL to 5 mL so
final [KI] is 0.500 x 1/5 = ?
volume H2O2 before mix is 4 mL
total volume after mix is 5 (4 from H2O2 + 1 from KI) so the H2O2 has been diluted from 4 to 5 mL.'
0.88 x 4/5 = ?
KI before mix is 0.500.
volume KI before mix is 1 mL.
total volume after mix is 5 (1 from KI + 4 from H2O2). The 0.500 M KI has been diluted from 1 mL to 5 mL so
final [KI] is 0.500 x 1/5 = ?
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