A 0.239 gram sample of a gas in a 100-ml flask exerts a pressure of 603 mm Hg at 14 Celsius. What is the gas? (Hint: don’t forget BRINCLHOF) A) chlorine

B) nitrogen

C) krypton

D) xenon

D) oxygen

Please help me and explain how u got ur answer

A) chlorine

Explanation:

To solve this question we can use:

PV = nRT

In order to solve the moles of the gas. With the moles and the mass we can find the molar mass of the gas to have an idea of its identy:

PV = nRT

PV / RT = n

Where P is pressure: 603mmHg * (1atm / 760mmHg) = 0.7934atm

V = 100mL = 0.100L

R is gas constant = 0.082atmL/molK

T is absolute temperature = 14°C + 273.15 = 287.15K

0.7934atm*0.100L / 0.082atmL/molK*287.15K = n

3.37x10⁻³ moles of the gas

In 0.239g. The molar mass is:

0.239g / 3.37x10⁻³ moles = 70.9g/mol

The gas with this molar mass is Chlorine, Cl₂:

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

im pretty sure its b