What pressure will be needed to lwave the vloume of 77.4l ay helium at 98.ok pa to a vloume of 60.0l

P₂ = 126.42 kpa

Explanation:

Given data:

Initial volume = 77.4 L

Initial pressure = 98.0 Kpa

Final volume = 60.0 L

Final pressure = ?

Solution:

Formula:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = Initial volume

P₂ = Final pressure

V₂ = Final volume

Now we will put the values in formula.

P₁V₁ = P₂V₂

98.0 kpa × 77.4 L  = P₂× 60.0 L

7585.2 Kpa.L = P₂× 60.0 L

P₂= 7585.2 Kpa.L/ 60.0 L

P₂ = 126.42 kpa

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