Question

(W) 36.0 cm (ii) - 59,0 cm The following problems are due next week but you should work on them with your group today 4.(20pts) Valves closed He ideal gas) (Flask on left) Neideal gas) (Flask in middle) Paskon Right V-3.50 L V = 6.85 L V=3.00 L P = 442 torr P=640 torr P=0 (vacuum) When the valves are open, the gases will mix. Assuming that the temperature is constant, what is the final total pressure in torr? Hint: First compute the partial pressures of each gas after it expands to fill the entire space after the valves are open 5.(25pts) The Ca in a sample of limestone weighing 0.142g was precipitated by using oxalate ions CO. to produce Ca C,04() via the reaction Ca (aq) + CO2 (aq) -----> CaCO(S) If the resulting Ca C,04() weighed 0.153 g, what was the % Ca in the original sample? Bonus Problem (+6pts. No partial credit. A 500.0 mg table of an antacid containing Mg(OH), Al(OH), and a binder was dissolved in 50.0 mL of 0.500 M HCI. The resulting solution was acidic, and it needed 30.9 mL of 0.255 M NaOH for neutralization. (a) Calculation the number of moles of OH ions in the tablet. (b) If the tablet contains 5.0% binder, how many milligrams of Mg(OH), and how many of Al(OH), does the tablet contain?

Answer 1

Q4) first calculate moles of He ( g ) and Ne ( g ) when valve is closed

n( He ) = (442 / 760) atm x (3.5 litre) / ( R litre- atm / mol K )( T K )

n( Ne ) = ( 640 / 760 ) atm x ( 6.85 litre) / ( R litre - atm / mol K )x T k

Final volume after opening valve = 3.5 lt + 6.85 lt + 3 lt = 13.35 litre

partial pressure after opening valve

p ( He ) = n( He ) RT / V ( total)

= ( 442 / 760 ) x 3.5 RT / (RT X 13.35) =0.15 atm

p ( Ne ) = n( Ne) RT / V ( total )

= ( 640 / 760 ) x 6.85 RT / (RT X 13.35 )= 0.43 atm

P ( total ) = P ( He ) + P ( Ne ) = 0.15 atm + 0.43 atm

= 0.58 atm.

P ( total ) = ( 0.58 atm ) x ( 760 torr ) / 1 atm = 440.8 torr