Answer:
V₂ =31.8 mL
Explanation:
Given data:
Initial volume of gas = 45 mL
Initial temperature = 135°C (135+273 =408 K)
Final temperature = 15°C (15+273 =288 K)
Final volume of gas = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 45 mL × 288 K / 408 k
V₂ = 12960 mL.K / 408 K
V₂ =31.8 mL
Answer:
4. trigonal planar symmetry.
Explanation:
The sp2 hybridization is formed with one s and two p atomic orbitals and form trigonal planar symmetry. In sp2 hybridization, there are same valence shell in both the orbitals and it gives three equivalent sp2 hybridized orbitals that are separated by 120 degrees giving trigonal planar symmetry.
Hence, the correct answer is "4. trigonal planar symmetry.".
The most concentrated solution is b
Almost all properties are common to elements within a single group on the periodic table. They react with water in the same way, they have the same number of valence electrons thereby having the same valency, the number of shells in the atom of the element increases by one as we move down the group.
In general, they have the same chemical properties as chemical properties depend on the number of electrons in the valence shell i.e. the outermost shell in the atom of an element.
Answer:
7. 3–ethyl–6 –methyldecane
8. 5–ethyl–2,2–dimethyl–4–propyl–4 –heptene
Explanation:
It is important to note that when naming organic compounds having two or more different substituent groups, we simply name them alphabetically.
The name of the compound given in the question above can be written as follow:
7. Obtaining the name of the compound.
Compound contains:
I. Decane.
II. 3–ethyl.
III. 6 –methyl.
Naming alphabetically, we have
3–ethyl–6 –methyldecane
8. Obtaining the name of the compound.
Compound contains:
I. 2,2–dimethyl.
II. 4–propyl.
III. 4 –heptene.
IV. 5–ethyl.
Naming alphabetically, we have
5–ethyl–2,2–dimethyl–4–propyl–4 –heptene
