Emissions of SO2 generated during the combustion of fossil fuels can be reduced<span> by treating the flue gases before they are emitted into the atmosphere via the stack; this is termed Flue Gas Desulphurisation (FGD). Flue gas desulphurisation systems can be classified as either Regenerable or Non-regenerable.
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Answer:
Volume required from standard solution = 4675 mL
Explanation:
Given data:
Final volume = 75.0 mL
Final molarity = 130 M
Molarity of standard solution = 2.000 M
Volume required from standard solution = ?
Solution:
We use the formula,
C₁V₁ = C₂V₂
here,
C₁ = Molarity of standard solution
V₁ = Volume required from standard solution
C₂ = Final molarity
V₂ = Final volume
Now we will put the values in formula,
C₁V₁ = C₂V₂
2.000 M × V₁ = 130 M × 75.0 mL
V₁ = 9750 M. mL / 2.000 M
V₁ = 4675 mL
Answer:
Fluorine > Selenium > Arsenic > Potassium > Argon
Explanation:
Electron affinity describes the ability or readiness or tendency of an atom to gain an electron.
The higher the value, the higher the tendency. Electron affinity depends on the on the nuclear charge and atomic radius. When nuclear charge is more, electron affinity is high, when atomic radius increases electron affinity reduces.
Noble gases such as Helium, Neon, and Argon would have 0 affinity for electrons because of their stable electronic configuration. From the list, Ar is the least in terms of electron affinity.
Potassium is a metal with large electropositivity which describes the tendency of an atom to lose electrons. Potassium would readily lose electrons instead of gaining.
Between Arsenic and Selenium: Arsenic belongs to group V and Selenium group VI. The two elements both belong to period IV on the periodic table. Across a period, electron affinity increases due to increase in nuclear charge. Therefore, Selenium would have a greater electron affinity compared to Arsenic.
Fluorine has the highest electron affinity of all. It needs just an electron to complete its octet.
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:
45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present
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