Answer: Hence, the equilibrium concentration of the
are 0.521, 0.632 and
respectively.
Explanation:
The balanced chemical reaction is:

At t = 0 0.522 0.633 0
At
0.522-x 0.633-2x 2x
The expression for
for the given reaction follows:
![K_c=\frac{[SO_2]^2}{[SO_2]\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSO_2%5D%5E2%7D%7B%5BSO_2%5D%5Ctimes%20%5BO_2%5D%7D)
We are given:

Putting values in above equation, we get:


Thus equilibrium concentration of
is = (0.522-x )= 
equilibrium concentration of
is = (0.633-2x )= 
equilibrium concentration of
is = (2x )= 
Answer:
tin
Explanation:
When copper is combined with another element, tin, it makes bronze, the first man made metal alloy
The equilibrium constant K₁ = Equilbrium constant K₂.
The equilibrium constant, K, of a reaction, is defined as:
"The ratio between concentration of products powered to their reaction quotient and concentration of reactants powered to thier reaction quotient".
For the reaction:
aA + bB ⇄ cC + dD
The equilibrium constant, K, is:
![K = \frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Now, assuming the reaction of the problem is 1:1:
A + B ⇄ C + D
![K = \frac{[C][D]}{[A][B]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%5BD%5D%7D%7B%5BA%5D%5BB%5D%7D)
The concentrations of the reactants are directly proportional to the volume added. Thus, we can assume that concentration = Volume. Replacing for K₁ and K₂:
![K_1 = \frac{[C][D]}{[10mL][10mL]} = K_1 = \frac{[C][D]}{100mL^2}](https://tex.z-dn.net/?f=K_1%20%3D%20%5Cfrac%7B%5BC%5D%5BD%5D%7D%7B%5B10mL%5D%5B10mL%5D%7D%20%3D%20K_1%20%3D%20%5Cfrac%7B%5BC%5D%5BD%5D%7D%7B100mL%5E2%7D)
In the same way:
![K_2 = \frac{[C][D]}{[1mL][100mL]} = K_2 = \frac{[C][D]}{100mL^2}](https://tex.z-dn.net/?f=K_2%20%3D%20%5Cfrac%7B%5BC%5D%5BD%5D%7D%7B%5B1mL%5D%5B100mL%5D%7D%20%3D%20K_2%20%3D%20%5Cfrac%7B%5BC%5D%5BD%5D%7D%7B100mL%5E2%7D)
Thus, we can say:
<h3>K₁ = K₂</h3><h3 />
Learn more about chemical equilibrium in:
brainly.com/question/4289021?referrer=searchResults
Answer:
2.3125g
Explanation:
Half-life referred to the time required for a quantity to reduce to half of its initial value, It used to calculate how unstable atoms undergo, or the period of time and atom can survive, radioactive decay.
Given:
t(1/2)= 9.25days
Initial mass of Thulium-167 = 48grams
We need to calculate the remaining amount after 37days.
Since we know that 1 half life = 9.25 days
Then 37 days means ( 37/9.25) half lives
37days means 4 half life
That means the 38grams of Thulium-167 will be halved by 4 times.
Then the ratio between the initial Amount and the amount remaining after 37 days can be calculated as. 0.5^(4)
= 37days × 0.5^(4)
= 2.3125g
the remaining amount of Thallium-167 after 37days is 2.3125g
There would be no more because everything had decrease to 0