You need 1.50 moles of ammonia (nh3) for a reaction at stp. what volume of ammonia must you measure?

also,

how many molecules of nacl are in 23.40 grams nacl?

1) 33.64 Liter will be volume of ammonia.

2) molecules of NaCl are in 23.40 grams NaCl.

Explanation:

1) Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = (at STP)

V = Volume of gas = ?

n = number of moles of gas = 1.50 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 273,15 K

Putting values in above equation, we get:

V = 33.64 L

2)

Where:

N = Number of particles / atoms/ molecules

n = Number of moles

= Avogadro's number

We have:

Molar mass of NaCl = 58.5 g/mol

n =

molecules of NaCl are in 23.40 grams NaCl.

the answer is the first option

explanation:

just googled it m8. not that hard

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