Answer:
The answer to your question is 0.79 moles of SnCl₂
Explanation:
Data
moles of SnCl₂ = ?
mass of FeCl₃ = 85.3 g
excess Sn
Balanced chemical reaction
2 FeCl₃ + 3 Sn ⇒ 3 SnCl₂ + 2 Fe
Process
1.- Convert the mass of FeCl₃ to moles
Molar mass of FeCl₃ = 56 + (35.5 x 3)
= 56 + 106.5
= 162.5 g
Use proportions to find the moles of FeCl₃
162.5 g -------------------- 1 mol
85.3 g ------------------- x
x = (85.3 x 1) / 162.5
x = 0.525 moles
2.- Find the number of moles SnCl₂
2 moles of FeCl₃ ----------------- 3 moles of SnCl₂
0.525 moles ----------------- x
x = (0.525 x 3) / 2
x = 0.79 moles of SnCl₂
Answer:
D
Explanation:
The picture depicts the data of the chemical, explaining that as the temperature rises, the chemical reaction rate would increase as well.
HNO₃ + H₂S → S + NO + H₂<span>O
Assign Oxidation Number:
L.H.S R.H.S
N in HNO</span>₃ = +5 +2 = N in NO
S in H₂S = -2 0 = S in S
Write Half cell Reactions:
Reduction Reaction:
3e⁻ + HNO₃ → NO -------(1)
Oxidation Reaction:
H₂S → S + 2e⁻ -------(2)
Multiply eq. 1 with 2 and eq. 2 with 3 to balance electrons.
6e⁻ + 2 HNO₃ → 2 NO
3 H₂S → 3 S + 6e⁻
Cancel e⁻s,
______________________________
2 HNO₃ + 3 H₂S → 2 NO + 3 S + H₂O
Balance Oxygen Atoms by multiplying H₂O with 4, Hydrogen atoms will automatically get balance.
2 HNO₃ + 3 H₂S → 2 NO + 3 S + 4H₂O
Answer:
The reaction is not at equilibrium and reaction must run in forward direction.
Explanation:
At the given interval, concentration of NO = 
Concentration of 
= 
Concentration of NOBr = 
Reaction quotient,
, for this reaction = ![\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%7D%7B%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D%7D)
species inside third bracket represents concentrations at the given interval.
So, 
So, the reaction is not at equilibrium.
As 
therefore reaction must run in forward direction to increase and make it equal to 
.
On adding salt.....The boiling temperature increases.....
So ∆t= KB * molality
=O.52*(58/58)/4
= O.52*1/4
= 0.13
So increase is 100+.13=100.13°c
