How many liters of Cl2 gas will you have if you are using 63 g of Na?

You will have 19.9L of Cl2

Explanation:

We can solve this question using:

PV = nRT; V = nRT/P

Where V is the volume of the gas

n the moles of Cl2

R is gas constant = 0.082atmL/molK

T is 273.15K assuming STP conditions

P is 1atm at STP

The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:

63g * (1mol / 70.906g) = 0.8885 moles

Replacing:

V = 0.8885mol*0.082atmL/molK*273.15K/1atm

V = You will have 19.9L of Cl2

can you attach a picture of the lettuce? : )

they are both synthetic elements

explanation:

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