Answer:
18.94%.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
- The integration law of a first order reaction is:
<em>kt = ln [A₀]/[A]</em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = 13,750 years.
[A₀] is the initial percentage of carbon-14 = 100.0 %.
[A] is the remaining percentage of carbon-14 = ??? %.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(13,750 years) = ln (100.0%)/[A]
1.664 = ln (100.0%)/[A]
Taking exponential for both sides:
5.279 = (100.0%)/[A]
<em>∴ [A]</em> = (100.0%)/5.279 = <em>18.94%.</em>
Answer:
First, we can test Solution 1. We know that Sodium Hydroxide is a strong base. If we test acids on blue litmus paper, they will turn red. If we test bases on red litmus paper, they will turn blue. So, you can test all the of the solutions- water, sodium hydroxide and hydrochloric acid with blue and red litmus paper. HCl, Hydrochloric acid is an acid, so it will turn blue litmus paper red. It will not turn red litmus blue. The acids will turn blue litmus paper red. The bases will turn red litmus paper blue. Only water is a neutral liquid, which will not turn blue litmus paper red or red litmus paper blue. It will not change the colour of it. Thus, if you test all the solutions with blue and red litmus paper, you will know which solution is water. Water is the only one which is neutral. It is the only solution which cannot change the colour of any litmus paper. Thus, you can identify it very easily.
Delta enthalpy = 2x386-3x1x432-3x942=-3350kJ/mol
Enzymes catalyze the chemical reactions, they act upon the reaction substrates and speed up the reaction. Enzymes have active sites, the places where the reaction substrates interact with the enzyme bringing about the conversion of substrates to products. So, as the enzyme concentration increases the rate of reaction increases till a point where the rate is leveled off. The rate does not further increase, as the substrate might have become limiting at that point. All the available amount of substrate would have been associated with the active sites of the enzymes. So, at that point although there is enough catalyst, lack of substrate would limit the rate of reaction.
Answer:
An Arrhenius Base
Explanation:
The definition of this is a base that is a hydroxide ion donor.
