Answer: The density of Ammonia is 0.648 g/l
Explanation:
Density = Mass/ Volume
Mass of one mole of Ammonia (NH3) = 17.031g
Volume =?
Using the ideal gas law we can determine the volume.
PV = nRT
P = 0.913 atm, V= ?, n = 1, R = 0.08206 L.atm/K, and T= 293K
Make V the subject of the formular, we then have;
V= nRT/ P = 1 mol x 0.08206 L.atm/ K.mol x 293 / 0.913 atm
V = 24.04358/ 0.913 = 26.3L
Having gotten the value of Volume in this question, we then go back to solve for density.
Density = Mass/ Volume
17.031g/ 26.3L = 0.64756 ≈ 0.648 g/l
Answer :]
to convert from g NaOH to mol NaOH. = 1.48 g NaOH are needed to neutralize the acid.
Depends on what the base is. You would reference the base dissociation chart for that value.
Base+salt > acid +alkali > neutralization i think this is the reaction
Volume of the tank is 5.5 litres.
Explanation:
mass of the CO2 is given 8.6 grams
Pressure of the gas is 89 Kilopascal which is 0.8762 atm
Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)
R = gas constant 0.0821 liter atmosphere per kelvin)
FROM THE IDEAL GAS LAW
PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)
no of moles = mass/atomic mass
= 8.6/44
= 0.195 moles
now putting the values in equation
V=nRT/P
= 0.195*0.0821*302/ 0.8762
= 5.5 litres.
As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.
