Answer: 4
Explanation:
Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n.
Azimuthal Quantum Number: This quantum number describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as 
. The value of this quantum number ranges from 
. When l = 2, the value of will be -2, -1, 0, +1, +2.
Given : a f subshell, thus l = 3 , Thus the subshells present would be 3, 2, 1, 0 and thus n will have a value of 4.
Also electrons give are 32.
The formula for number of electrons is 
.
Thus principal quantum no will be n= 4.
Answer:
Nitrifying Bacteria are a group of aerobic bacteria important in the nitrogen cycle as converters of soil ammonia to nitrates, compounds usable by plants. An example is nitrosomonas or nitrobacter and species in that family.
The schematic diagram is attached below, which summarises the oxidation of ammonia or free nitrogen in the soil to nitrates for the cowpea plant's utilisation.
Answer is 4 ½ billion.
Scientists think the earth's age is 4 ½ billion. This calculation is based on the radioactive dating. Since the formation of the earth, the time period has been divided into 3 major geological time periods as eon. They are phanerozoic, proterozoic and archean.
Answer:
293.1 mL.
Explanation:
- Boyle's law states that: at a constant temperature the pressure of a given mass of an ideal gas is inversely proportional to its volume.
- It can be expressed as: <em>P₁V₁ = P₂V₂,</em>
P₁ = 546.0 mm Hg, V₁ = 350.0 mL.
P₂ = 652.0 mm Hg, V₂ = ??? mL.
<em>∴ V₂ = (P₁V₁)/(P₂)</em> = (546.0 mm Hg)(350.0 mL) / (652.0 mm Hg) = <em>293.1 mL.</em>
Answer:
Hello my Friend! The answer is: Manganese(III) oxide is a transition metal compound. The oxidation state of manganese in this compound is +3 , and the chemical formula of the compound is Mn2O3.
Explanation:
Manganese can have two oxidation states: +2 and +3, but in this case, the "(III)" indicates that in this compound, the state of oxidation is +3.
