Answer:
1) SO₄
²⁻ : (+6)
H₂S : (-2)
Explanation:
a) <u>Sulfate reducers</u> are widespread in muds and other sediments, water-logged soils, etc., environments that contain SO₄ ²⁻ and become anoxic as a result of microbial decomposition.
Sulfate (SO₄ ²⁻), the most oxidized form of sulfur (+6), <u>is reduced</u> by these
sulfate-reducing bacteria. The end product of sulfate reduction is hydrogen sulfide, H₂S, (oxidation number -2) an important natural product that participates in many biogeochemical processes. The H₂S they generate is responsible for the pungent smell (like that of rotten eggs) often encountered near coastal ecosystems. When sulfate-reducing bacteria grow, the H₂S formed from SO₄ ²⁻ reduction combines with the ferrous iron to form black, insoluble ferrous sulfide, which is not toxic. This is important for the conservation of the environment.
b) The net ionic equation under acidic conditions is:
4 H₂ + SO₄²⁻ + H⁺ → HS⁻ + 4 H₂O
Global reaction: SO₄²⁻ + 2H⁺ → H₂S + O₂
Answer: Option (e) is the correct answer.
Explanation:
Formula to calculate radius is as follows.
p(r) = 
= 
= 0
+ 
= 0
= 
r = 
Thus, we can conclude that most likely radius at which the electron would be found is .
Answer: The pH of the solution is 11.2
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in ml
moles of 
= 
(1g=1000mg)
Now put all the given values in the formula of molality, we get
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
According to stoichiometry,
1 mole of gives 2 mole of 
Thus 0.0298 moles of gives =
moles of
Putting in the values:
![pOH=-\log[0.0596]=2.82](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B0.0596%5D%3D2.82)
Thus the pH of the solution is 11.2
To get the ∆S of the reaction, we simply have to add the ∆S of the reactants and the ∆S of the products. Then, we get the difference between the ∆S of the products and the ∆S of the products. If the <span>∆S is negative, then the reaction spontaneous. If the otherwise, the reaction is not spontaneous.</span>
