Answer:
105 grams PbI₂
Explanation:
Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)
moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole
moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant
moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂
grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)
Answer:
4.5 M
Explanation:
70.0 ml was mixed in 3.00 M of Na2SO4
30.0 ml was mixed in 1.00 M of NACL
The first step is to convert 70 ml to liters
= 70/1000
= 0.07 liters
The formular for molarity is
moles/liters
The number of moles in Na2S04 can be calculated as follows
Let y represent the number of moles
3M= y moles/0.07
= 3×0.07
= 0.21 moles
Since Na2So4 has 2 moles of Na then the number of moles is
= 2×0.21
= 0.42 moles
Convert 30ml to liters
= 30/1000
= 0.03 liters
The number of moles in Nacl can be calculated as follows
Let y represent the number of moles
1M= y moles/0.03
= 1×0.03
= 0.03 moles
Since Nacl has 1 mole of Na then the number of moles is
= 1 × 0.03
= 0.03 moles
Therefore the final Na+ can be calculated as follows
Total moles = 0.03 moles + 0.42 moles
= 0.45 moles
Total liters= 0.07 liters + 0.03 liters
= 0.1 liters
Na+ = 0.45/0.1
= 4.5 M
Hence the final Na+ in the solution is 4.5 M
Answer:
Passivation of Oxide layers of the metals.
Explanation:
Passivation is a non-electrolytic finishing process that makes most metals rust-resistant. The prosses removes free iron from the surface by using either nitric or citric acid. When this happens, it results to an inert, protective oxide layer that is very slow or less likely to chemically react with air and cause corrosion.
Passivity caused many of the metals several minutes to begin to react. Once the finishing process that makes metals less likely to react was eroded, reaction was initiated vigorously.
The given question is incomplete, here is a complete question.
At 400 K, this Reaction has 
What Is 
at 400 K for the following reaction?
(A) 8.2 x 10⁻⁴
(B) 2.9 x 10⁻²
(C) 6.7 x 10⁻⁷
(D) 1.6 x 10⁻⁷
Answer : The correct option is, (C) 
Explanation :
The given chemical equation follows:
The equilibrium constant for the above equation, .
We need to calculate the equilibrium constant for the following equation of above chemical equation, which is:
, 
The equilibrium constant for the doubled reaction will be the square of the initial reaction.
Or, we can say that
If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.
The value of equilibrium constant for the following reaction is:
Hence, the value of equilibrium constant for the following reaction is,
