Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘c.

This is an incomplete question, here is a complete question.

Consider the reaction between nitrogen and oxygen gas from dinitrogen monoxide.

Given: delta Hrxn= +163.2kJ

Calculate the entropy change in the surroundings when this reaction occurs at 25 degrees C.

Answer : The entropy change in the surroundings is, -547.6 J/K

Explanation :

Formula used to calculate the entropy change in the surroundings is:

where,

= entropy change in the surrounding

Q = heat energy

T = temperature =

Given:

Now put all the given values in the above formula, we get:

Therefore, the entropy change in the surroundings is, -547.6 J/K

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.