Explanation:
a) Using the provided information about the density of gold, the sample size, thickness, and the following equations and comersion factors, find the area of the gold leaf:

Gold 

First, find the volume of the sample and then find the area of the sample.


b. Using the provided information from part
), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :


In this question, you are given the NaOH volume but asked for concentration.
Don't forget that for every 1 mol of NaOH there will be 1 mol OH- ion, but for every 1 mol of H2SO4 there will be 2 mol of H- ion.
To neutralize you need the same amount of OH- and H+, so the equation should be:
OH-= H+
<span>35.50cm3 * x*1= 25cm3* 0.2mol/dm3 *2
</span>x= 10/35.5 mol/dm3= 0.2816/dm3
First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
Q is unlike K value it describes the reaction that is not at equilibrium.
by considering this reaction:
aA+ bB⇄ cC
and our reaction is:
Br2 + Cl2 ⇄ 2 BrCl
According to Q low:
Q= concentration of products/concentration of reactants
but this equation in the gaseous or aqueous states only.
∴ Q = [BrCl]^2 / [Br2] [Cl2]
and we have [Br2] = 0.00366 m [Cl2]= 0.000672 m [BrCl] = 0.00415 m
by substitution:
= [0.00415]^2 / ( [0.00366] * [0.000672])
∴ Q = 7