How many liters of gaseous hydrogen bromide at 58°c and 0.891 atm will a chemist need if she wishes to prepare 3.50 l of 1.20 m hydrobromic acid?

We need 128 liters of a gaseous hydrogen bromide

Explanation:

Step 1:

Temperature = 58°C

Pressure = 0.891 atm

Volume ofHydrobromic acid we want to prepare is 3.50 L (1.20M)

Step 2: Calculate moles HBr

Moles HBr = molarity * volule

Moles HBr = 1.20 * 3.50 L

Moles HBr = 4.20 moles HBr

Step 3: Calculate volume of Hydrogen bromide

pV = nRT

V=(nRT)/P

⇒ with V = the volume of gaseous hydrogen bromide = ?

⇒with p = the pressure of the gas = 0.891 atm

⇒ with n = the number of moles of hydrogen bromide = 4.2 moles

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

⇒ with T = the temperature = 58.0 °C = 273 + 58 = 331 Kelvin

V = (4.2*0.08206*331)/0.891

V = 128 L

We need 128 liters of a gaseous hydrogen bromide

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.