Answer:
can you help mine please
How many molecules of chlorine are needed to react with 56.Og of iron to form Iron (III) chloride (FeCl3)?
Answer:
the answer is 6
Explanation:
there is 3 hydrogen molecules in NH3 and there's 2 molecules of NH3 so in total, there are 6 hydrogen molecules on the products side.
Density = mass/volume = 2000/4000 = 0.5 grams/cm3. Hope this hopes!
Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
The heat absorbed by the water is
Q = 500 (4.18) (32.2 - 25)
Q = 15048 J
The enthalpy of fusion of the sodium acetate is:
<span>ΔHf = Q / m
</span><span>ΔHf = 15048 / 100
</span>ΔHf = 150.48 J/g
