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kirza4 [7]
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.)
Chemistry
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

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What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M
Fed [463]

<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>

What is benzoic acid found in?

  • Some natural sources of benzoic acid include: ​Fruits:​ Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
  • ​Spices:​ Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 =  0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

pH = pKa + log\frac{base}{acid}

4.2 + log\frac{0.05}{0.045} = 4.245

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

pH = pKa + log\frac{base}{acid}

4.245 = 3.75 + log\frac{base}{acid}

log\frac{base}{acid} = 0.5

\frac{base}{acid} = 3.162

Now we must solve the equation above. This will be done using the following values

\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

brainly.com/question/24052816

#SPJ4

3 0
2 years ago
What is the molarity of a 9.13 L soda that contains 13.83 of sugar?
choli [55]

Answer:

1.52 M

Explanation:

Molarity of a solution is calculated as follows:

Molarity = number of moles (n) ÷ volume (V)

Based on the information given in this question,

Volume of soda (V) = 9.13 L

number of moles = 13.83 mol

Molarity = 13.83 ÷ 9.13

Molarity = 1.52 M

7 0
3 years ago
32.33 mL of 1.031 M potassium hydroxide were required to reach the endpoint of a titration of 50.00 mL of nitric acid. What was
tester [92]

The molar concentration of the nitric acid solution was 0.6666 mol/L.

<em>Balanced equation</em>: KOH + HNO_3 → KNO_3 + H_2O

<em>Moles of KOH</em>: 32.33 mL KOH × (1.031 mmol KOH /1 mL KOH)

= 33.33 mmol KOH

<em>Moles of HNO_3</em>: 33.33 mmol KOH× (1 mmol HNO_3/1 mmol KOH)

= 33.33 mmol HNO_3

<em>Concentration of KOH</em>: <em>c </em>= "moles"/"litres" = 33.33 mmol/50.00 mL

= 0.6666 mol/L

4 0
3 years ago
The difference between the initial (first) position and the final position of an object
Anna [14]
It is called DISPLACEMENT of the object....
3 0
3 years ago
A precipitation reaction is caused by mixing 100. mL of 0.25 M K2Cr2O7 solution with 100. mL of 0.25 M Pb(NO3)2 solution. When t
Sergeeva-Olga [200]

Answer:

Neither is affected

Explanation:

The reaction occurs as follows:

K₂Cr₂O₇ + Pb(NO₃)₂ → PbCr₂O₇ + 2K⁺ + 2NO₃⁻

That means per mole of reaction you will have two moles of both K⁺ and NO₃⁻.

But volume is also doubled, doing that concentration of spectator ions doesn't change.

Right answer: Neither is affected

I hope it helps!

8 0
3 years ago
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