Ask question

Ask question

All categories

3 years ago

7

As with any drug, aspirin (acetylsalicylic acid) must remain in the bloodstream long enough to be effective. Assume that the rem

oval of aspirin from the bloodstream into the urine is a first-order reaction, with a half-life of about 3 hours. The instructions on an aspirin bottle say to take 1 or 2 tablets every 4 hours. If a person takes 1 aspirin tablet, how much aspirin remains in the bloodstream when it is time for the second dose? (A standard tablet contains 400. mg of aspirin.)

1 answer:

Olenka [21]3 years ago

4 0

Answer:

180 mg

Explanation:

For a first-order reaction, we can calculate the amount of aspirine (A) at a certain time (t) using the following expression.

$A=A_{0}.e^{-k.t}$

where,

k: rate constant

A₀: initial amount

If we know the half-life ( $t_{1/2}$ ) we can calculate the rate constant.

$k=\frac{ln2}{t_{1/2}} =\frac{ln2}{3h} =0.2h^{-1}$

When t = 4 h and A₀ = 400 mg, A is:

$A=400mg.e^{-0.2h^{-1}\times 4h} =180mg$

You might be interested in

How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?

Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

S1=50%

V2=250ml

S2= 20%

We all know that

V1S1=V2S2

∴V1= V2×S2÷S1

∴V1= V2S2×1/S1

∴V1= 250×20÷50

∴V1= 100ml

6 0

3 years ago

What is the mole ratio between oxygen and carbon dioxide?<br><br> C25H52 + 38 O2 → 25 CO2 + 26 H2O

user100 [1]

Answer:

38 : 25

Explanation:

First thing's first, we have to confirm if the reaction is indeed balanced.

The equation of the reaction is given as;

C25H52 + 38 O2 → 25 CO2 + 26 H2O

From the reaction, 38 moles of O2 produces 25 moles of CO2

The ratio is given as;

38 : 25

4 0

3 years ago

In the compound sodium methoxide (naoch3), there is ________ bonding

zvonat [6]

Maybe covalent bonding?

5 0

4 years ago

Why pie bond are not perticipate in hybridaization

DedPeter [7]

The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.

3 0

3 years ago

Consider the following reaction between mercury(II) chloride and oxalate ion:

siniylev [52]

Answer : The reaction rate will be, $1.9\times 10^{-4}M/s$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2HgCl_2(aq)+C_2O_2^{4-}(aq)\rightarrow 2Cl^-(aq)+2CO_2(g)+HgCl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_2^{4-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_2^{4-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{k(0.164)^a(0.45)^b}{k(0.164)^a(0.15)^b}\\\\9=3^b\\(3)^2=3^b\\b=2$

Dividing 3 from 2, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{k(0.164)^a(0.45)^b}{k(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

Now, calculating the value of 'k' by using any expression.

Putting values in above rate law, we get:

$3.2\times 10^{-5}=k(0.164)^1(0.15)^2$

$k=8.7\times 10^{-3}M^{-2}s^{-1}$

Now we have to determine the reaction rate when the concentration of $HgCl_2$ is 0.135 M and that of $C_2O_2^{-4}$ is 0.40 M.

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

$\text{Rate}=(8.7\times 10^{-3})\times (0.135)^1\times (0.40)^2$

$\text{Rate}=1.9\times 10^{-4}M/s$

Therefore, the reaction rate will be, $1.9\times 10^{-4}M/s$

6 0

3 years ago