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Taya2010 [7]
3 years ago
14

Many drugs decompose in blood by a first-order process. two tablets of aspirin supply 0.60 g of the active compound. after 30 mi

n, this compound reaches a maximum concentration of 2 mg/100 ml of blood. if the half-life for its breakdown is 90 min, what is its concentration (in mg/100 ml) 4.0 h after it reaches its maximum concentration?
Chemistry
2 answers:
Lunna [17]3 years ago
4 0

The concentration of drug after 4.5 hoursis \boxed{{\text{0}}{\text{.25 mg/100 mL}}}.

Further Explanation:

Radioactive decayis responsible to stabilizeunstable atomic nucleusaccompanied by the release of energy.

Half-life is the duration after which half of the original sample has been decayed and half is left behind. It is represented by {t_{{\text{1/2}}}}.

The relation between rate constant and half-life period for first-order reaction is as follows:

k = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}                                                                                                    …… (1)

Here,

{t_{{\text{1/2}}}} is half-life period.

k is rate constant.

Substitute 90 min for {t_{{\text{1/2}}}} in equation (1).

\begin{aligned}k &= \frac{{0.693}}{{90{\text{ min}}}} \\&= 7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}} \\\end{aligned}  

Radioactive decay formula is as follows:

{\text{A}} = {{\text{A}}_0}{e^{ - kt}}                                        …… (2)

Here

A is the concentration of sample after time t.

t is the time taken.

{{\text{A}}_0} is the initial concentration of sample.

<em>k</em> is the rate constant.

The time for the process is to be converted into min. The conversion factor for this is,

1{\text{ hr}} = 60\min  

Therefore time taken can be calculated as follows:

\begin{aligned}t&= \left( {4.5{\text{ hr}}} \right)\left( {\frac{{60{\text{ min}}}}{{1{\text{ hr}}}}} \right) \\&= 270{\text{ min}} \\\end{aligned}  

Substitute 2 mg/100 mL for {{\text{A}}_0} , 7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}} for k and 270 min for t in equation (2).

\begin{aligned}{\text{A}} &= \left( {\frac{{2{\text{ mg}}}}{{100{\text{ mL}}}}} \right){e^{ - \left( {7.7 \times {{10}^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}} \right)\left( {270{\text{ min}}} \right)}} \\&= \frac{{0.25{\text{ mg}}}}{{100{\text{ mL}}}} \\\end{aligned}  

Learn more:

  1. What nuclide will be produced in the given reaction? brainly.com/question/3433940
  2. Calculate the nuclear binding energy: brainly.com/question/5822604

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Radioactivity

Keywords: half-life, 0.25 mg/100 mL, 2 mg/100 mL, A, k, t, rate constant, 4.5 h, 270 min, radioactive decay formula.

Firdavs [7]3 years ago
3 0
<span>The half-life of a first-order reaction is determined as follows: 

</span>t½<span>=ln2/k

From the equation, we can calculate the </span><span>first-order rate constant:

</span>k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³

When we know the value of k we can then calculate concentration with the equation:

A₀ = 2 g/100 mL 

t = 2.5 h = 150min

A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml

   = 6.3 × 10⁻⁴ mg / 100ml


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Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4
inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

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3 years ago
Semi-conductors have properties of both metals and non metals.which are two examples of metalloids
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The six metalloids are boron, silicon, germanium, arsenic, antimony, and tellerium.
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Are isotopes similar to ions, yes or no?
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Yes

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How many moles of mercury (II) oxide are needed to produce 65.00 g of mercury?
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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
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Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

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let's plug in the values in the above equation and calculate heat gained by combined system.

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q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


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