Answer:
3–ethyl–4–methylhexane.
Explanation:
To name the above compound, do the following:
1. Determine the functional group of the compound.
2. Locate the longest continuous carbon chain. This gives the parent name of the compound.
3. Identify the substituent group attached to the compound.
4. Give the substituent the lowest possible count.
5. Combine the above to name the compound.
Now, we shall name the compound given in the question above as follow:
1. The compound contains only single bond. Therefore, the compound belong to the alkane family.
2. The longest continuous carbon chain is 6 i.e hexane.
3. The substituent group attached are:
i. Methyl, CH3.
ii. Ethyl, CH2CH3.
4. we shall name the substituents alphabetically i.e ethly will come before methyl. Therefore,
Ethyl is located at carbon 3.
Methy is located at carbon 4.
5. Therefore, the name of the compound is:
3–ethyl–4–methylhexane.
The appropriate response is gamma radiation. Alpha particles can be halted via air. UV radiation can be halted by a typical layer of clothing.Beta particles can be ceased by the thick plastic suit. Just gamma radiation can enter the substantial suit. It must be halted by thick dividers of lead or cement.
Answer:
Darker colored items absorb more of the sunlight.
Explanation:
Lighter colored item tend to reflect more of the sunlight
B, because water is changing its state from solid to liquid (it's fusion in portuguese, don't know in english), so while it's changing, water has 2 states at the same time.
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
