Answer:
0.1066 hours
Explanation:
A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.
t1/2 = ln2/k
t1/2 = ln2/6.5 h⁻¹
t1/2 = 0.1066 h
The half-life of the pesticide is 0.1066 hours.
Answer: b
explanation: i had the question on a test and got it right.
The length in kilometers of a row of 4.34 x 1023 hydrogen atoms is 4.34 x 10¹⁴ km.
<h3>
Length of the entire hydrogen atoms</h3>
The length of the entire hydrogen atom is calculated as follows;
Length of the row = number x diameter of one
Length of the row = 4.34 x 10²³ x 10⁶ x 10⁻¹²
Length of the row = 4.34 x 10¹⁷ m
Length of the row = 4.34 x 10¹⁴ km
Thus, the length in kilometers of a row of 4.34 x 1023 hydrogen atoms is 4.34 x 10¹⁴ km.
Learn more about diameter of hydrogen atom here: brainly.com/question/13796082
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Answer:
Is better use the Benedict's test by the increase in the amount of the products if the enzyme is a reductase
Explanation:
The Benedict's test works by the reaction of the reducing sugars with the ion cupric of the reactive. If the enzyme is a reductase (degrades polysaccharides into bi o monosaccharides), it should cut the polysaccharide bond and the products would react with the Benedict's cupric ion
I hope you undestand me
