The equilibrium constant for this reaction at 350°C is D. 282.
<h3>Equilibrium constant</h3>
A dynamic chemical system approaches chemical equilibrium constant when enough time has passed and its composition no longer exhibits any discernible propensity to change further. The equilibrium constant of a chemical reaction is the value of its reaction quotient in this condition. The equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture for a specific set of reaction conditions. Understanding equilibrium constants is crucial for comprehending many chemical systems as well as biological processes like the transport of oxygen by hemoglobin in the blood and the maintenance of acid-base homeostasis in the human body. There are many different kinds of equilibrium constants, including stability constants, formation constants, binding constants, association constants, and dissociation constants.
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A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.
Br2(g) + I2(g) ↔ 2IBr(g)
When the mixture has come to equilibrium, the concentration of iodine monobromide is
1.190 M. What is the equilibrium constant for this reaction at 350°C? Show step-by step explanation.
A) 3.55 × 10^3
B) 1.24
C) 1.47
D) 282
E) 325
This compound is Boron selenate. Molar mass of B2(SeO4)3 is 450.4948 g/mol.
Electrons, everything is pretty much based around the likeliness of electrons to be swapped or shared between atoms
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,
A(t) = (A(o))(0.5)^(t/h)
where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,
0.89 = (0.5)(t / 5730 years)
The value of t from the equation is 963.34 years.
<em>Answer: 963 years</em>
