Answer:
Kindly check the attachment for the diagram representing the cell and showing all necessary components including: anode, cathode, electron flow, cation flow and anion flow.
Explanation:
So, the reaction in the concentration cell is given below as;
Cu2+(1.00 M) → Cu2+(0.0100 M).--(1).
The anode = Cu2+(0.0100 M) because it has lesser Concentration, thus, lesser potential value.
Cathode = Cu2+(1.00 M) because it has higher Concentration, hence higher potential value.
It must be noted that in the digaram depicting the Reaction, the electrons moves from the anode part of the cell to the cathode part of the cell and this is done through an external circuit. The following are the things that happens at each electrode;
At the Anode: in here is where oxidation occurs and Cu^2+ is released into the solution.
At the cathode: in here is where the reduction occur and the Cu^2+ moves in the direction to where the Cu electrode is, thus, causing the deposition of Cu.
Answer:
(a) A strong acid has a greater tendency to lose its protons.
(b) A strong acid has a higher Ka than a weak acid.
(c) A strong acid has a lower pKa than a weak acid.
Explanation:
A strong acid has the ability to completely transfer their protons to the water, making a complete dissociation. Instead, a weak acid only dissociates partially, how much it dissociates depends on the acid dissociation constant (Ka).
The weak acids always are in equilibrium, and the equilibrium depends on the acid dissociation constant.
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Thus, a stronger acid with a bigger Ka produces more dissociation and a higher concentration of protons.
The equation that defines pKa is:
Therefore, a higher pKa means a lower Ka and also a weaker acid.
Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
Answer:
Explanation:
25.8 ml of .095 N NaOH is needed to neutralise the remaining acid
equivalent of NaOH used = 25.8 x .095 / 1000 = .002451 gm equivalent .
acid remaining = .002451 gm equivalent .
acid initially taken = 100 ml of .1 N / 1000 = . 01 gm equivalent
acid reacted with metal = .01 -.002451 = .007549 gm equivalent
This must have reacted with same gram equivalent of metal oxide
.007549 gm equivalent = .15 gm of metal oxide
1 gm equivalent = 19.87 gm
equivalent weight of metal = 19.87 - equivalent weight of oxygen
= 19.87 - 8 = 11.87 .
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