Answer:
Density, melting point. and magnetic properties
Explanation:
I can think of three ways.
1. Density
The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.
It should be possible to distinguish these even with high school equipment.
2. Melting point
Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).
A Bunsen burner can easily reach these temperatures.
3. Magnetic properties
You can use a Gouy balance to measure the magnetic susceptibilities.
In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.
In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.
A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.
A sample of CuS will be attracted by the magnetic field and show an increase in weight.
In the picture below, you can see the sample partially suspended between the poles of an electromagnet.
I believe that the answer is 11.5
Explanation:
It is known that
of
=
.
(a) Relation between
and
is as follows.

Putting the values into the above formula as follows.

= 
= 3.347
Also, relation between pH and
is as follows.
pH = ![pK_{a} + log\frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
= 
= 3.44
Therefore, pH of the buffer is 3.44.
(b) No. of moles of HCl added = 
=
= 0.0116 mol
In the given reaction,
will react with
to form 
Hence, before the reaction:
No. of moles of
= 
= 0.15 mol
And, no. of moles of
= 
= 0.12 mol
On the other hand, after the reaction :
No. of moles of
= moles present initially - moles added
= (0.15 - 0.0116) mol
= 0.1384 mol
Moles of
= moles present initially + moles added
= (0.12 + 0.0116) mol
= 0.1316 mol
As,
= 

= 
= 3.347
Since, volume is both in numerator and denominator, we can use mol instead of concentration.
As, pH = ![pK_{a} + log \frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
= 3.347+ log {0.1384/0.1316}
= 3.369
= 3.37 (approx)
Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.