Answer:
5 L.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 10 L
Initial pressure (P1) = 2.5 atm
Final pressure (P2) = 5 atm
Final volume (V2) =.?
Since the temperature is constant, we shall apply the Boyle's law equation to determine the new volume of the gas. This can be obtained as follow:
P1V1 = P2V2
2.5 × 10 = 5 × V2
25 = 5 × V2
Divide both side by 5
V2 =25/5
V2 = 5 L
Thus, the new volume of the gas is 5 L
Answer:-
People with diabetes must monitor their blood glucose levels constantly.
Blood glucose levels are nowadays measured by the help of enzyme glucose oxidase. The enzyme glucose oxidase catalyzes the oxidation of glucose to gluconic acid.
By this way all of the glucose is measured at one time. At the same time hydrogen peroxide is produced.
The hydrogen peroxide reacts with a second color producing chemical. The concentration of the glucose can be related to the intensity of color produced
First we have to understand what a half life is. A half life is a measure of time when the amount of a certain object is 50% of the original amount. Hence the answer for this is letter A. The initial amount is 100. fifty percent of 100 is 50 and that happens after 15 hours. Hence, 15 hours is the half life period.
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
