Answer: 
Explanation:
Entropy is the measure of randomness or disorder of a system.
A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.
1. 
As 4 moles of gaseous reactants are changing to 2 moles of gaseous products, the randomness is decreasing and the entropy is negative
2.
As 9 moles of gaseous reactants are changing to 10 moles of gaseous products, the randomness is increasing and the entropy is positive.
3.
As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.
4.
As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive
5. 
As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.
The correct name for the hydrocarbon would be option 2. 2 - methyl - 2 - pentene.
Answer:
202 g/mol
Explanation:
Let's consider the neutralization between a generic monoprotic acid and KOH.
HA + KOH → KA + H₂O
The moles of KOH that reacted are:
0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol
The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.
1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:
0.2688 g/1.33 × 10⁻³ mol = 202 g/mol
Answer:
I think its C a scientific journal article
Explanation:
Any kind of journal is considered a primary source because journals contain info that the original author wrote. Encyclopedias are considered teritary sources, but im not sure if that counts ...so id go with journals.
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
