<span>Molarity is expressed as the number of moles of solute per volume
of the solution while molality is expressed as the number of moles solute per
mass of solution. We calculate as follows:</span>
5.74 mol / kg (1.238 kg/L) = 7.10612 mol / L or 7.11 M
Answer:B is right
Explanation: Al loses electrons and is oxidised, Au gains electron. And is reduced
Answer:
The answer to your question is CuSO₄
Explanation:
To answer your question just remember the following information
- A chemical reaction is divided into sections
reactants and products
reactants on the left side of the reaction
products on the right side of the reaction
- All the symbols have a meaning
(s) means that that compound is in solid phase
(aq) means that that compound is dissolved in solution.
Then, the answer is CuSO₄
Answer: option C. Copper (II) chloride
Explanation:
To name CuCl2, we need to know the oxidation state of Cu in the compound as chlorine always have oxidation on —1 in all its compound. The oxidation state of Cu can be calculated as follows:
Cu + 2Cl = 0 (since the compound has no charge)
Cl = —1
Cu + 2(—1) = 0
Cu —2 = 0
Collect like terms
Cu = 0 +2
Cu = +2
Therefore, the oxidation state of Cu in CuCl2 is +2.
The name of the compound will be copper(ii) chloride, since cupper has oxidation state +2 in the compound.
<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
