How many moles of chloroform, CHCl3, are required to fill a 253-mL flask at 100.0C and 940 torr?

0.0103 mole

Explanation:

Data obtained from the question include:

V (volume) = 253mL

Recall: 1000mL = 1L

Therefore, 253mL = 253/1000 = 0.253L

T (temperature) = 100°C = 100 + 273 = 373K

P (pressure) = 940 torr

Recall: 760torr = 1atm

Therefore, 940torr = 940/760 = 1.24atm

R (gas constant) = 0.082atm.L/Kmol

n (number of mole) =?

Using the ideal gas equation PV = nRT, the number of mole 'n' can be obtained as follow:

PV = nRT

1.24 x 0.253 = n x 0.082 x 373

Divide both side by 0.082 x 373

n = (1.24 x 0.253)/(0.082 x 373)

n = 0.0103 mole

Therefore, the number of mole of chloroform, CHCl3, required to fill the flask is 0.0103 mole

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: