Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
V = nRT/P
V = (2.1 moles Cl2) (0.08203 L - atm / mol - K) (273K) / (1 atm)
V = 47 L</span>
Answer:
a. The second run will be faster.
d. The second run has twice the surface area.
Explanation:
The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

The area of the 10.0 cm³-sphere is:

The area of each 1.25 cm³-sphere is:

The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²
The ratio of 8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00
Since the surface area is doubled, the second run will be faster.
When sulfate (SO₄²⁻) serves as the electron acceptor at the end of a respiratory electron transport chain, the product is hydrogen sulfide (H₂S).
How sulfate acts as electon acceptor and electron donor?
- Sulfate (SO₄²⁻) is used as the electron acceptor in sulfate reduction, which results in the production of hydrogen sulfide (H2S) as a metabolic byproduct.
- Many Gram negative bacteria identified in the -Proteobacteria use sulfate reduction, which is a rather energy-poor process.
- Gram-positive organisms connected to Desulfotomaculum or the archaeon Archaeoglobus also utilise it.
- Electron donors are needed for sulfate reduction, such as hydrogen gas or the carbon molecules lactate and pyruvate (organotrophic reducers) (lithotrophic reducers).
Learn more about the Electron transport chain with the help of the given link:
brainly.com/question/24372542
#SPJ4
When an electron moves from a higher energy shell to a lower one, energy in the form of light, more specifically a wavelength of a photon is released.
Answer:
Far right
Explanation:
The qualities indicated here are of the element Sulphur (S) and any element of same qualities i.e color pale yellow, nature solid, electrical conductivity none, should be placed at far right corner of the periodic table or position 3 mentioned in the attached picture.
Hope it help!