<span>Boyle's Law is k = PV so
Initial k = 13.0 L x 4.0 atm = 52 L atm
Final kf = 6.5 L x 8 atm = 52 L atm
The gas obeys Boyle's Law
The answer with two significant figures separated by a comma is k = 52, kf = 52.</span>
Answer : The reagent present in excess and remains unreacted is, 
Solution : Given,
Moles of 
= 3.00 mole
Moles of = 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 3.00 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the reagent present in excess and remains unreacted is,
Bases
A base is a substance that dissociates into more hydroxide ions (-OH-) when dissolved in water. Bases are also good proton acceptors. Bases, therefore, reduce the number of H+ and increase OH- hence raising the pH of the solution.
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)
Explanation:
Other properties of bases is that they are bitter to the taste and they feel slippery when touched. Strong bases are nonthlese very corrosive like acids. Bases turn red litmus paper blue. Most alkali hydroxides such as NaOH are bases.
Learn More:
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Carbonated drinks are a mixture of many compounds such as glucose(sugar) and others, while water is simply its own molecule.
Answer : Option 4) Region of the most probable electron location.
Explanation : As per the electron cloud model of the atom, an orbital is a region where the probability of finding an electron is highest. According to this model which was used to identify the probable location of the electrons when they go around the nucleus of an atom.
This electron cloud model was different from the older Bohr atomic model by Niels Bohr.
