<u>Answer:</u> The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane
<u>Explanation:</u>
The IUPAC nomenclature of alkanes are given as follows:
- Select the longest possible carbon chain.
- For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
- A suffix '-ane' is added at the end of the name.
- If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.
We are given:
An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms
The image of the given alkane is shown in the image below.
Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane
Answer:
42 days
Explanation:
Half life = 8.4 days
Starting mass = 40.0 g
Time = ?
Final Mass = 1/16 * 40 = 2.5 g
First Half life;
Remaining mass = 40 / 2 = 20g
Second Half life;
Remaining mass = 20 / 2 = 10g
Third Half life;
Remaining mass = 10 / 2 = 5g
Fourth Half life;
Remaining mass = 5 / 2 = 2.5g
Time = Number of half lives * Duration of half life = 5 * 8.4 = 42 days
orange juice is a liquid.
Answer:
Atomic number
Explanation:
Because atomic number is unique for each element and never changes
Answer:
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺
Explanation:
This is an easy redox reaction:
Br₂ + 2KI → I₂ + 2KBr
We determine the oxidation states.
0 for the elements at ground state.
K does not change the oxidation state during the reaction.
Bromine is reduced to bromide (oxidation state decreases)
and iodide is oxidized to Iodine (oxidation state increases)
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
In oxidation, electrons are released while in reduction, the electrons are gained. To make the ionic equation, we just add K⁺
So we sum both reactions
Br₂ + 2e⁻ + 2I⁻ + 2K⁺ ⇄ 2e⁻ + I₂ + 2Br⁻ + 2K⁺
We cancel the electrons on both sides of the equation:
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺