Would it be lack of water and food?
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
Question:
<em>What effects does the concentration of reactants have on the rate of a reaction?</em>
Answer:
<em>Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.</em>
<em>Increasing the concentration of reactants generally increases the rate of reaction because more of the reacting molecules or ions are present to form the reaction products. ... When concentrations are already high, a limit is often reached where increasing the concentration has little effect on the rate of reaction.</em>
Hope this helps, have a good day. c;
Answer:
A.) 1.845
Explanation:
You can find the pOH using the following equation:
pOH = -log[OH⁻]
Since NaOH dissociates into 1 Na⁺ and 1 OH⁻, the concentration of both ions is 0.0143 M.
pOH = -log[OH⁻]
pOH = -log[0.0143]
pOH = 1.845
Volume percent<span> or </span>volume/volume percent<span> (v/v%) is used when preparing solutions of liquids. It will have units of volume of the smaller composition substance over the volume of the solution. We calculate as follows:
12.5 mL ethanol = .225 mL ethanol / 1 mL solution ( V )
V = 55.56 mL of the 22.5 % by volume ethanol solution is needed
Hope this answers the question.</span>
