Dividing through by A(#1s) and simplifying we obtain (#2x)| | A(#2s) #2s A mixtu

Question

Dividing through by A(#1s) and simplifying we obtain (#2x)| | A(#2s) #2s A mixture) - #1x #1s A( #1s #1s A mixture A#1s) #1x Since the relationship follows the form: y -mx+ b, then a plot of versus #2s AHl will give a line with a slope of #2xl and an intercept of casi That is, the concentration of the unknown component in the mixture #2x, equals the slope times the concentration of the standard solution for component #2. Likewise, the concentration of the unknown component in the mixture #lx, equals the intercept times the concentration of the standard solution for component #1, or simply Let's apply this method to the analysis of the metals in a United States five-cent coin. 0.8 Copper 0.6 Nickel Coin Nickel 0.4 0.2 0.0 400 500 600 700 800 Wavelength (nm) Figure 1-Absorbance spectra of nickel, copper and five-cent coin

Explanation / Answer

Since the equation of straight line corresponding to the third graph is given we may follow the calculations aa below

Given y=mx+c

Where m is the slope and y is intercept

Here equation is 6.5707x-6.2574

Ourcomponent1 is Nickel

So unknown composition of 1 in mixture equals intercept times the concentration of standard nickel solution

=( 0.6524)(0.2M) =0.13048M Ni

Similarly concentration of component 2 ( copper) is slope times the concentration of standard copper

=( 6.5707).(O.1M)= 0.65707M Cu

This is the method of calculations of two unknowns given in the procedure. But for further calculation of percentage composition total mass of the coin is necessary which is a missing data and hence the percentage composition cannot be computed

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