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liberstina [14]
2 years ago
12

A 90 lb person is prescribed a drug in the amount of 50mg/kg body per day to be delivered in 2 doses per day. Each tablet contai

ns 250mg of the medicine. How many tabletes should be given per dose?
Chemistry
1 answer:
geniusboy [140]2 years ago
7 0

Answer:

4 tabletes/dose

Explanation:

mass person = (90 Lb)×(Kg/2.20462 Lb) = 40.823 Kg body

amount of medicine = 50 mg/Kg body.day

frecuency = 2 dose/day

⇒ amount of medicine/day =  (50 mg/kg body.day)×(40.823 Kg body) = 2041.168 mg/day

⇒ mg medicine/dose = (20141.168 mg/day)×(day/ 2 dose) = 1020.584 mg medicine/dose

∴ mg medicine/tablet = 250 mg medicine/tablet

⇒ # tabletes/dose = (1020.584 mg medicine/dose)×(tablet/250 mg medicine) = 4.082 tabletes/dose

⇒ # tabletes/dose ≅ 4 tabletes/dose

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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
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<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.
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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

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