Answer:
Drought, resulting in a 65% reduction in water levels.
Answer:
a) Limiting: sulfur. Excess: aluminium.
b) 1.56g Al₂S₃.
c) 0.72g Al
Explanation:
Hello,
In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.
b) By stoichiometry, the produced grams of aluminium sulfide are:

c) The leftover is computed as follows:

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.
Best regards.
Answer:
oxidation-reduction or redox reaction
Explanation:
n
Answer:
0.00050553
Explanation:
when the power of ten is negative, move the decimal to the left
hope this helped!
Answer:
The answer to your question is P2 = 170.9 torr
Explanation:
Data
Volume 1 = 12.1 l Volume 2 = 21.1 l
Temperature 1 = 241 °K Temperature 2 = 298°K
Pressure 1 = 546 torr Pressure 2 = ?
Process
To solve this problem use the combined gas law.
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = T1V1T2 / T1V2
-Substitution
P2 = (241 x 12.1 x 298) / (241 x 21.1)
-Simplification
P2 = 868997.8 / 5085.1
-Result
P2 = 170.9 torr