M₁=6.584 g
m₂=4,194 g
m(H₂O)=m₁-m₂
w(H₂O)=m(H₂O)/m₁
w(H₂O)=(m₁-m₂)/m₁
w(H₂O)=(6.584-4.194)/6.584=0.3630 (36.30%)
the percentage by mass of water in the hydrate 36.30
Answer:
The volume of the system will change and maybe the shape of the system boundary
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:
the answer is c. their atomic masses are different clearly because an atom of gold has 79 protons and the atom can be divided multiple times. An atom of silver has an atomic number of 47. 47 electrons. Clearly different. Hope it helps :)
Explanation:
