Answer:
Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq)
1.35 V
Explanation:
Given that the reduction potential of
Cu^2+(aq) + e- -----> Cu^+(aq) is +0.15 V
While the reduction potential of
Au3+(aq) + 3 e- -----> Au(s) is + 1.50V
It is clear that the Cu^+(aq)/Cu^2+(aq) system is the anode while Au^3+(aq)/Au(s) system is the cathode based on the reduction potentials shown above. The number of electrons transferred (n) =3
E°cell = E°cathode - E°anode
E°cell= 1.50-0.15
E°cell= 1.35 V
Moles (mol) = mass (g) / molar mass (g/mol)
Mass of NaCl = 21.7 g
Molar mass of NaCl = <span>58.4 g/mol
Hence, moles of NaCl = </span>21.7 g / 58.4 g/mol = 0.372 mol
Hence moles of NaCl in the mixture is 0.372 mol.
Let's assume that mixture has only given compounds and free of impurities.
Then, we can present this as a mole percentage.
mole % = (moles of desired substance / Total moles of the mixture) x 100%
Hence,
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
Total moles of mixture = moles of NaCl + KCl + LiCl
Mass of KCl = 3.74 g
Molar mass of NaCl = 74.6 g/mol
Hence, moles of NaCl = 3.74 g / 74.6 g/mol = 0.050 mol
Mass of NaCl = <span>9.76 g
</span>Molar mass of NaCl = 42.4 g/mol
Hence, moles of NaCl = 9.76 g / 42.4 g/mol = 0.230 mol
Total moles = 0.372 mol + 0.050 mol + 0.230 mol = 0.652 mol
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
= (0.372 mol / 0.652 mol) x 100%
= 57.06%
Hence, mixture has 57.06% of NaCl as the mole percentage.
3, 9, 16 are correct! 17 should be b because buffered means that it resists change in pH
The correct answer is (B) Adding a dilute solution of HCl
<u>EXPLANATION</u>
The presence of carbonate ions can be tested by adding a dilute acid to the solution. The acid displaces Carbon (IV) oxide from the solution. Using HCl, and a carbonate of metal X.
XCO₃₍s₎ + 2HCl₍aq₎⇒ XCl₂₍aq₎+ H₂O₍l₎ + CO₂₍g₎
The gas produced is tested using calcium hydroxide to confirm whether it is carbon (IV) oxide.
