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
Maybe her old shoes had soft worn out bottoms and she slips in them. So her new shoes had more grip than her old ones so they kept her from falling.
C. quadruples the rate
<h3>Further explanation</h3>
Given
The rate law :
R=k[A]²
Required
The rate
Solution
There are several factors that influence reaction kinetics :
- 1. Concentration
- 2. Surface area
- 3. Temperature
- 4. Catalyst
- 5. Pressure
- 6. Stirring
The rate is proportional to the concentration.
If the concentration increased, the reaction rate will increase
The reaction is second-order overall(The exponent is 2)
The concentration of A is doubled, the reaction rate will increase :
r = k[A]² ⇒ r= k[2A]²⇒r=4k[A]²
<em>The reaction rate will quadruple.</em>
C = vf
c stands for the speed of waves (which is a constant that is 3 x 10^8)
v stands for the wavelength (which is given)
f stands for frequency (what we are solving for)
3 x 10^8 = (1.08 x 10^-6)f
Divide both sides by the given wavelength
f = 2.78 * 10^14 seconds
The given molarity of sodium hydroxide solution = 2.0 M
The required concentration of sodium hydroxide is 65 mL of 0.6 M NaOH
Converting 65 mL to L:
Calculating the moles of NaOH in the final solution:
Finding out the volume of 2.0 M solution taken to prepare the final solution:
Therefore, 19.5 mL of 2.0 M NaOH solution and make it up to 65 mL to prepare 0.6 M NaOH solution.
