A mixture of hydrogen and argon gases is maintained in a 6.47 L flask at a pressure of 3.43 atm and a temperature of 85 °C. If the gas mixture contains 1.10 grams of hydrogen, the number of grams of argon in the mixture is g.

The mixture contains 8.23 g of Ar

Explanation:

Let's solve this with the Ideal Gases Law

Total pressure of a mixture = (Total moles . R . T) / V

We convert T° from °C to K → 85°C + 273 = 358K

3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L

(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles

0.756= Total moles from the mixture

Moles of Ar + Moles of H₂ = 0.756 moles

Moles of Ar + 1.10 g / 2g/mol = 0.756 moles

Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206

We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g

The mass of argon is 8.42 grams

Explanation:

Step 1: Data given

Volume = 6.47 L

Pressure = 3.43 atm

Temperature = 85 °C = 358 K

Mass of hydrgen = 1.0 grams

Step 2: Calculate the number of moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the number of moles = TO BE DETERMINED

⇒with p = the pressure = 3.43 atm

⇒with V = the volume = 6.47 L

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

⇒with T =the temperature = 85 °C  = 358 K

n = (3.43*6.47) / (0.08206 * 358)

n = 0.7554 moles

Step 3: Calculate moles Ar

Thus,  0.7554 = mol of Ar + mol of H2

There is 1.10 grams ofH2 gas

Moles H2 =  1.10 grams / 2.02 g/mol

Moles H2 = 0.5446 moles H2

Moles Ar = 0.7554 - 0.5446 = 0.2108 moles

Step 4: Calculate mass Ar

Mass Ar = moles Ar * molar mass Ar

Mass Ar = 0.2108 moles *39.95 g/mol

MAss Ar = 8.42 grams

The mass of argon is 8.42 grams

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)