A 1.00 L flask is filled with 1.20 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.300 atm . What is the partial pressure of argon, PAr, in the flask?

Partial pressure of Ar = 0.73 atm

Explanation:

Given data:

Volume of flask = 1.00 L

Mass of Ar = 1.20 g

Temperature = 25 °C (25+273.15 K= 298.15 k)

Total pressure = 1.300 atm

Partial pressure of Ar = ?

Solution:

First of all we will calculate the number of moles of Ar:

Number of moles = mass/ molar mass

Number of moles = 1.20 g/ 39.95 g/mol

Number of moles = 0.03 mol

Pressure of Ar:

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

P×1.00 L =  0.03 mol ×  0.0821 atm.L /mol.K× 298.15 k

P = 0.73 atm.L /1.00 L

P = 0.73 atm

Total pressure = Partial Pressure of Ar + Partial pressure of ethane

1.300 atm = 0.73 atm + P(ethane)

P(ethane) = 1.300 atm - 0.73 atm

P(ethane) = 0.57 atm

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:

i dont know

explanation: