Answer:
The answer is C.
Explanation:
Mass is converted to the energy binding a nucleus together.
Explanation:
Entropy means the amount of randomness present within the molecules of the body of a substance.
Relation between entropy and microstate is as follows.
S = 
where, S = entropy
= Boltzmann constant
= number of microstates
This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.
Also, let us assume that
= 1, and
= 0.833
Therefore, change in entropy will be calculated as follows.

= 
= 
= 
or, = 
Thus, we can conclude that the entropy change for a particle in the given system is
J/K particle.
Answer:
The amount of CaCl2 produced depends on the amount of HCl in the reaction.
Explanation:
The amount of HCl is used completelyin the reaction unlike CaCO3 which remains after reaction.
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.
You forgot to attach the text
BTW reduction is the action or fact of making a specified thing smaller or less in amount, degree, or size.