32. A solution suspected of containing copper is analyzed. To determine the standard curve, the absorbance of five standards ranging from 0.125 mg/L Cu to 1.25 mg/L Cu was measured and graphed against the concentration. The slope of the linear best-fit line was determined to be 0.6136 and the y-intercept was 0.0142. If three samples of the unknown solution have absorbance values of 0.438, 0.434, and 0.439 what is the concentration of the unknown

0.689 mg/L Cu

Explanation:

The equation of the best-fit line is:

y = 0.6136x + 0.0142

Where y is the absorbance and x is the concentration in mg/L.

Now using that equation we calculate x when y is equal to 0.438, 0.434 & 0.439.

1) 0.438 = 0.6136x + 0.0142

x = 0.691 mg/L

2) 0.434 = 0.6136x + 0.0142

x = 0.684 mg/L

3) 0.439 = 0.6136x + 0.0142

x = 0.692 mg/L

Finally we calculate the mean of the three concentrations:

the answer is c.the valence of the atoms combine .