Lead(II) nitrate will react with iron(III) chloride to produce the precipitate lead(II) chloride as shown in the balanced reaction
2FeCl3(aq) + 3Pb(NO3)2(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Calculating the amount of the precipitate lead(II) chloride each reactant will produce:
mol PbCl2 = 0.050L Pb(NO3)2 (0.100mol/1L)(3mol PbCl2/3mol Pb(NO3)2)
= 0.00500mol PbCl2
mol PbCl2 = 0.050L FeCl3 (0.100mol FeCl3/1L)(3mol PbCl2/2mol FeCl3) = 0.00750mol PbCl2
The reactant Pb(NO3)2 produces a lesser amount of the precipitate PbCl2, therefore, the lead(II) nitrate is the limiting reagent for this reaction.
Answer:0.8742j/g°C
Explanation: SOLUTION
GIVEN
length of bar=1.25m
mass 382g
temperature= 20°C to 288°C
Q=89300J
Specific Heat Capacity will be calculated using
Q=mC∆T
where
C = specific heat capacity
Q = heat
m = mass
Δ T = change in temperature
C=Q/ m∆T
=89300/382X(288-20.6)
=0.8742j/g°C
9.184 liters CH2O at STP
I think this is correct. Good luck
I think it would be C.100.5cm or D.100.5ml hope that helps
