If 4.50 g of methane gas (CH4) is in a 2.00-L container at 35°C, what is the pressure in the
container?

The answer to your question is Pressure = 3.55 atm

Explanation:

Data

R = 0.082 atm L/mol°K

mass = 4.5 g

volume = 2 L

temperature = 35°C

Pressure = ?

Process

1.- Calculate the moles of Methane

molar weight = 12 + 4 = 16 g

                          16 g of Methane 1 mol

                          4.5 g                     x

                           x = (4.5 x 1)/16

                           x = 0.281 moles

2.- Convert temperature to Kelvin

°K = 273 + 35

    = 308

3.- Use the ideal gas law to solve this problem

          PV = nRT

- Solve for P

           P = nRT / V

- Substitution

           P = (0.281)(0.082)(308) / 2

- Simplification

           P = 7.097 / 2

- Result

            P = 3.55 atm

The pressure in the container is 3.55 atm

Explanation:

Step 1: Data given

Mass of methane gas = 4.50 grams

Volume = 2.00 L

Temperature = 35 °C = 308 K

Molar mass CH4 = 16.04 g/mol

Step 2: Calculate moles CH4

Moles CH4 = mass CH4 / molar mass CH4

Moles CH4 = 4.50 grams / 16.04 g/mol

Moles CH4 = 0.281 moles

Step 3: Calculate pressure

p*V = n*R*T

⇒with p = the pressure in the container = TO BE DETERMINED

⇒with V = the volume of the container = 2.00L

⇒with n = the moles of methane = 0.281 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ T = the temperature = 308 K

p = (n*R*T)/V

p = (0.281 * 0.08206 * 308 ) / 2.00

p = 3.55 atm

The pressure in the container is 3.55 atm

it takes on an electronical energy

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3