Answer:
Explanation:
Hello,
For the given chemical reaction:
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
Finally, we compute the percent yield with the obtained 2.10 g:
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Answer:
Anode:
3Mg(s) ----------> 3Mg2+(aq) + 6e
Cathode:
2Al3+(aq) +6e ---------> 2Al(s)
Explanation:
Anode:
3Mg(s) ----------> 3Mg2+(aq) + 6e
Cathode:
2Al3+(aq) +6e ---------> 2Al(s)
Magnesium is more electro positive than aluminum hence it functions as the anode. Six electrons are lost/gained in the redox process as shown in the oxidation and reduction half reaction equations above. Magnesium is oxidized to magnesium ion while aluminum is reduced to elemental aluminum.
Answer: The average kinetic energy is proportional to the absolute temperature of the gas. Gas particles are in random motion. Gas particles have no volume. And the collisions between gas particles are elastic.
Explanation: I heard this question before, I think. Hope this helps!
Answer : The amount of heat released is, -121.04 KJ
Solution : Given,
Enthalpy of reaction, 
= -890 KJ
Mass of methane = 6 g
Molar mass of methane = 44 g/mole
First we have to calculate the moles of methane.
The given balanced reaction is,
From the balanced reaction we conclude that
1 mole of methane releases heat = -890 KJ
0.136 moles of methane releases heat = 
Therefore, the amount of heat released is, -121.04 KJ
