Suppose that you wanted to determine the density of a small aystal to confirm that is silicon. From the literature, you know silicon a density 2.33 /cm^ 3 would you prepare of the liquid mixture having that density from pure samples of CHGl 3 (d bullet1,492 g/mL) and 2.890g/mL) ? ( Note: 1mL = 1 cm ^ 3

Volume of CHCl3 required, V(CHCl3) = 8.02 mL

Volume of CHBr3 required, V(CHBr3) = 11.98 mL

Explanation:

Note: The complete question is incomplete. The complete question is given below:

Suppose now that you wanted to determine the density of a small yellow crystal to confirm that it is silicon. From the literature, you know that silicon has a density of 2.33 g/cm3. How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of CHCl3 (d = 1.492 g/mL) and CHBr3 (d = 2.890 g/mL)? (Note: 1 mL = 1 cm3.)

V(CHCl3) = mL

V(CHBr3) = mL

Density of required mixture of CHCl3 and CHBr3 equals the density of silicon = 2.33 g/mL

Density of CHCl3 = 1.492 g/mL

Density of CHBr3 = 2.890 g/mL

Let the fraction of CHCl3 required be y

Fraction of CHBr3 required will be 1 - y

Density of mixture = ( fraction of CHCl3 * density of CHCl3) + (fraction of CHBr3 * density of CHBr3)

2.33 = (y * 1.492) + (1 - y) * 2.890

2.33 = 1.492y + 2.890 - 2.890y

-0.56 = -1.398y

y = 0.401

Therefore, volume of CHCl3 required = 0.401 * 20 mL = 8.02 mL

Volume of CHBr3 required = (20 - 8.02) = 11.98 mL

b. examine and describe cells

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