Answer:
C. A stretch of raised land, often built to help control flooding and the flow of a river.
Explanation:
Embankments are made typically of construction adjacent to the river from spoil excavated from its bank.
Answer:
B) -4.1 units
Explanation:
According to this question, a state property X has a value 89.6 units. It undergoes the certain changes as follows:
- first increase by 3.6 units
- then increase by another 18.7 units
- then decrease by 12.2 units
- and finally attains a value of 85.5 units
This can be mathematically represented by 89.6 - {3.6 + 18.7 - 12.2 - x) = 85.5
To get x, we say;
89.6 + 3.6 = 93.2
93.2 + 18.7 = 111.9
111.9 - 12.2 = 99.7
99.7 - 85.5 = 14.2units.
The changes that occured is represented as follows:
= (3.6 + 18.7) - (12.2 + 14.2)
= 22.3 - 26.4
= -4.1 units
Atoms are never rearranged
Answer:
Reagent A: PBr₃
Reagent B: Mg in Et₂O.
Explanation:
Hello,
In this case, your facing a problem in which a carboxylic acid is produced starting by an alcohol. More specifically, cyclopentanol must react with phosphorous tribromide in order to yield bromocyclopentane which is more likely to produce a carboxylic acid, therefore, reagent A is PBr₃.
On the other hand, by means of the production of the specified product, bromocyclopentane must react with carbon dioxide and magnesium in diethyl ether in acidic media to promote the production of the cyclopentanoic acid via the grignard reaction (substitution of the bromine by the carboxyle group), therefore, reagent B is Mg in Et₂O.
Best regards.
Answer:
16.6 mg
Explanation:
Step 1: Calculate the rate constant (k) for Iodine-131 decay
We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.
k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹
Step 2: Calculate the mass of iodine after 8.52 days
Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.
ln I = ln I₀ - k × t
ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day
I = 16.6 mg
