A sample of nitrogen gas is collected in a 100ml container at a pressure of 688 mmhg and a temperature of 565 degrees Celsius how many grams of nitrogen gas are presented in this sample?

Mass = 36.4 ×10⁻³ g

Explanation:

Given data:

Volume of container = 100 mL ( 100/1000 = 0.1 L)

Pressure = 688 mmHg (0.9 atm)

Temperature = 565°C ( 565 + 273 = 838 K)

Mass in grams of nitrogen = ?

Solution:

According to ideal gas equation,

PV = nRT

n = number of moles

P = pressure

R = general gas constant

T = temperature

From this equation we will determine the number of moles and then we we will calculate the mass.

n = PV/RT

n = 0.9 atm × 0.1 L / 0.0821 atm. L / mol.K × 838 K

n = 0.09 / 68.8

n = 0.0013 mol

or

n = 1.3×10⁻³ mol

Mass of nitrogen gas:

Mass = number of moles × molar mass

Mass = 1.3×10⁻³ mol ×  28 g/mol

Mass = 36.4 ×10⁻³ g

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:

it's filtration, but this is an assumption so my apologies if i got this