261.976 which rounds to 261.98(if rounding to the hundredths.
Answer:
Co2 means carbon oxygen 2 (carbon dioxide and oxygen combined
Answer:
Benzene: 37.5 Torr
Methylbenzene: 12.5 Torr
Explanation:
By Raoult's Law, each substance in a gas mixture contributes to the total pressure of the mixture proportionally to their respective mole fraction. So,
Ppartial = x*P°
Where x is the mole fraction (0.5 for each one because it's equimolar), and P° is the vapor pressure.
Benzene: Ppartial = 0.5 * 75 = 37.5 Torr
Methylbenzene: Ppartial = 0.5 * 25 = 12.5 Torr
Answer: This option is incorrect: <span>B. Covalent compounds are held together by much stronger interparticle forces than are ionic compounds.
Justification:
Ionic bonds, held by ionic compounds, are much stronger than covalent bonds, held by covalent compounds.
In ionic bonds one element yields one or more electrons forming a cation (a positively charged ion) and the other element accepts the electrons forming an anion (a negatively charged ion).
The anion and the cation are electrostatically atracted by each other. This electrostatic atraction force, named ionic bond, is very strong.
As result of this, the ionic compounds form strong crystals with high boiling and fusion points. A good example of this the sodium chloride, formed by the union of cation Na(+) and anion Cl(-).
The covalent bonds are result of sharing electrons and do not form ions. This bond is weaker than the ionic bond.
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Given what we know, we can confirm that if further increases in substrate concentration do not result in further increases in reaction rate, then an enzyme is likely saturated.
<h3>What does it mean for an enzyme to be saturated?</h3>
Enzymes work by binding to the substrate in specific zones of the enzyme. The zones are known as the active sites on enzymes. Since enzymes have a limited amount of these zones, once they are all bonded to a substrate, we can say that it is saturated.
Therefore, the saturation of enzymes allows us to explain how further increases in substrate concentration do not result in further increases in reaction rate.
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