Hi! Could someone please explain how to answer these questions? Thanks!
1. 5.00 g of magnesium carbonate has reacted with 25.0 mL of 0.500 M HCl. The volume of CO2 produced at 0 degrees C and 100 kPa is closest to:
A) 0.142 L
B) 0.155 L
C) 0.284 L
D) 1.35 L
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Hey!

The first thing you should always do is write out the chemical equation.
}+\ce{2HCl(aq)}\rightarrow\ce{MgCl2(aq)}+\ce{H2O(l)}+\ce{CO2(g)})
This requires you to find the limiting reagent (the one that runs out first).
So you find the moles of magnesium carbonate used by
dividing mass by molar mass.
5 / 84.31 = 0.0593molAnd you find the moles of the acid by
multiplying the concentration by the volume.
0.5 x 0.025 = 0.0125mol^This is the limiting reagent since it has less moles than the other reactant.
So there will be 0.00625mol of CO
2 because of the 1:2 mole ratio.
To find volume of gas,
multiply the moles of gas by 22.71L at 0'C0.00625 x 22.71 = 0.142LThus the answer is A.

For the galvanic cell, you'd refer to the reduction table for the nickel potential (-0.25 negative means it prefers oxidation over reduction).
Chlorine's potential is (+1.4 positive means it prefers reduction over oxidation).
Anode: Oxidation
Cathode: ReductionThus, the anode is nickel, and the cathode is chlorine gas at the inert platinum electrode.
It will flow from nickel to chlorine because it always flows from anode to cathode across the external wire.
}+\ce{Cl2(g)}\rightarrow\ce{Ni^{2+}(aq)}+\ce{2Cl^{-}(aq)})
The chlorine ions in solution do not produce a gas because it does not lose electrons, but nickel does.
The E* value is just nickel plus chlorine (remember to multiply nickel by -1 since it undergoes oxidation.
E*=0.25+1.4=1.65V (spontaneous reaction coz it's positive)The colour change observed would be that the anode cell (nickel) becomes more green as more ions of nickel are produced.
Hope this helps
