TNotice that, since the slope value in your equation represents y/x, the units o

Question

TNotice that, since the slope value in your equation represents y/x, the units of your slope would be sec/mL. We are trying to measure the rate of a reaction, so we need this flipped (mL/sec). So, in this case, the reaction rate is not your slope of the line, but rather then inverse slope. In other words, given the general line equation y = mx + b, you the inverse of the slope is 1/m. Determine the inverse of the slope (the reaction rate) for each of your trend lines, and record them. Make a data table in either Word or Excel for Table 5 and include the rates, as well as the information below. Show your work in a calculations section below your table.

Trial 1.

Trial 2.

Trial 3.

Please Show all calculations.

Table 5: Rate Reaction Rates for H2O2 and IKI

Trial

H2O2

IKI

RATE

1

0.88M

0.60M

2

0.88M

0.30M

3

0.66M

0.60M

Calculations (show work for calculating each rate):

Table 5: Rate Reaction Rates for H2O2 and IKI

Trial

H2O2

IKI

RATE

1

0.88M

0.60M

2

0.88M

0.30M

3

0.66M

0.60M

14.669x 23.256 R2 9738

Explanation / Answer

For Trial 1 trendline is

y = 14.669 x - 23.256

it is of form y =mx + c

given that rate is inverse of slope

here slope m = 14.669

so rate = 1/m = 1/14.669 = 0.0682 ml/sec Answer

For Trial 2 trendline is

y = 23.31 x + 18.202

it is of form y =mx + c

given that rate is inverse of slope

here slope m = 23.31

so rate = 1/m = 1/23.31 = 0.0429 ml/sec Answer

For Trial 3 trendline is

y = 19.02 x - 26.769

it is of form y =mx + c

given that rate is inverse of slope

here slope m = 119.02

so rate = 1/m = 1/19.02 = 0.0526 ml/sec Answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.