Volume = 22.4 dm3
n = 2 mol of H2
n = 1 mol of N2
Temperature = 273.15
All H2 reacts
reaction
N2 + 3H2 = 2NH3
1:3 ratio
Calculation:
N2 initial - N2 reacted = Final N2
1 - 2*(1/3) = 0.3333 mol of N2 left
H2 = 0 left
NH3 formed = 2/3*1 = 2/3 = 0.666
Total mol:
0.3333 + 0.666 = 1 mol
Apply the equation :
PV = nRT
P = nRT/V = 1*0.0082*(273.15)/(22.4) = 0.0999924 atm
PH2 = 0
PN2 = 1/3*0.0999924 = 0.0333308 atm
PNH3 = 2/3*0.0999924 = 0.0666616 atm
Answer is 0.0666616 atm
Answer:
Halogen / salt-former
Explanation:
Bromine is classified as an element in the 'Halogens' section which can be located in group 7 of the Periodic Table. The term "halogen" means "salt-former" and compounds containing halogens are called "salts".
Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.