Answer:
Kc for this equilibrium is 2.30*10⁻⁶
Explanation:
Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.
Being:
aA + bB ⇔ cC + dD
the equilibrium constant Kc is defined as:
![Kc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%2A%5BD%5D%5E%7Bd%7D%20%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.
In this case, being:
2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)
the equilibrium constant Kc is:
![Kc=\frac{[N_{2} ]*[H_{2} ]^{3} }{[NH_{3} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BN_%7B2%7D%20%5D%2A%5BH_%7B2%7D%20%5D%5E%7B3%7D%20%20%7D%7B%5BNH_%7B3%7D%20%5D%5E%7B2%7D%20%7D)
Being:
- [N₂]= 0.0551 M
- [H₂]= 0.0183 M
- [NH₃]= 0.383 M
and replacing:
you get:
Kc= 2.30*10⁻⁶
<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>
This question is missing the part that actually asks the question. The questions that are asked are as follows:
(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.
(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.
We can use the equation for a first order rate law to find the amount of material remaining after 4 days:
[A] = [A]₀e^(-kt)
[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.
(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.
4 days x 1 year/365 days = 0.0110
A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg
The decay of americium is so slow that no noticeable change occurs over 4 days.
(b) We can simply plug in the information of iodine-125 and solve for A:
A = (1.00)e^(-0.011 x 4)
A = 0.957 mg
Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.
Anti-acids because they contain Alkaline ion that chemically neutralizes stomach acid
<u>Answer:</u> The reaction proceeds in the forward direction
<u>Explanation:</u>
For the given chemical equation:
Relation of 
is given by the formula:
where,
= equilibrium constant in terms of partial pressure = ?
= equilibrium constant in terms of concentration = 
R = Gas constant = 
T = temperature = ![35^oC=[35+273]K=308K](https://tex.z-dn.net/?f=35%5EoC%3D%5B35%2B273%5DK%3D308K)
= change in number of moles of gas particles = 
Putting values in above equation, we get:
is the constant of a certain reaction at equilibrium while 
is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
The expression of for above equation follows:
We are given:
Putting values in above equation, we get:
We are given:
There are 3 conditions:
- When
; the reaction is product favored. - When
; the reaction is reactant favored. - When
; the reaction is in equilibrium
As, 
, the reaction will be favoring product side.
Hence, the reaction proceeds in the forward direction
Oxidation state of Pb in PbO2 is +4.
Oxidation state of Pb in PbCl2 is +2.
Oxidation state of Pb in Pb2O is +1.
Oxidation state of Pb in Pb4O3 is +6/4.
Hence option A. PbO2 is correct.
Hope this helps, have a nice day!
