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!!

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

the dissolution of polymer can be thermodynamically described in 2 steps.