750mL of gas is at STP. The pressure on the gas drops and the gas expands to 2L. How much did the pressure drop by

-0.625 atm

Explanation:

We need both volumes to be the same unit... s

750 mL = 0.750 L

And STP means the pressure is at 1 atm

V1 = 0.750 L

V2 = 2.00 L

P1 = 1 atm

And we will use this equation (Boyle's law) to get where we need to go:

P1*V1 = P2*V2

And we plug in everything:

(1 atm)*(0.750 L) = P2*(2.00 L)

0.750 L*atm = P2*(2.00 L)

[divide by 2.00 L on both sides, making L cancel out and leave us with atm as the units]

P2 = 0.375 atm

BUT!! We are not done yet!

The question asked the CHANGE in pressure, not the final pressure. S

P2 - P1 = change in P

0.375 - 1 = -0.625 atm

Change in P = -0.625 atm

Hopefully I did that right!!

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:

answer; i do believe the correct answer is (river); /// good luck