Answer:
Cl⁻ was oxidized.
Explanation:
- 4HCl + MnO₂ → Cl₂ + 2H₂O + MnCl₂
Oxidation can be defined as the process in which the oxidation number of a substance increases.
On the left side of the equation, Cl has a charge of -1 (in HCl); while on the right side of the equation Cl has a charge of 0 in Cl₂.
Thus, Cl⁻ was oxidized.
<u>Answer:</u> The equilibrium concentration of 
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of = 2.00 M
The given chemical equation follows:
<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of is 0.332 M
Given:
Half life(t^ 1/2) :30 years
A0( initial mass of the substance): 200 mg.
Now we know that
A= A0/ [2 ^ (t/√t)]
Where A is the mass that remains after t years.
A0 is the initial mass
t is the time
t^1/2 is the half life
Substituting the given values in the above equation we get
A= [200/ 2^(t/30) ] mg
Thus the mass remaining after t years is [200/ 2^(t/30) ] mg
Answer is: 9,7 L is needed to store helium gas.
n(He) = 0,80 mol.
p(He) = 204,6 kPa.
T = 300 K.
R = 8,314 J/K·mol; universal gas constant.
Use ideal law eqaution: p·V = n·R·T.
V = n·R·T / p.
V(He) = 0,80 mol · 8,314 J/K·mol · 300 K ÷ 204,6 kPa.
V(He) = 9,75 L.
