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liberstina [14]
3 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]3 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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A textbook measures 250 mm long, 225 mm wide and 50 mm thick. What is the volume of this book in mm3? What is the volume of this
Dvinal [7]

Answer:

2.81 × 10⁶ mm³

2.81 × 10⁻³ m³

Explanation:

Step 1: Given data

Length (l): 250 mm

Width (w): 225 mm

Thickness (t): 50 mm

Step 2: Calculate the volume of the textbook

The book is a cuboid so we can find its volume (V) using the following expression.

V = l × w × t = 250 mm × 225 mm × 50 mm = 2.81 × 10⁶ mm³

Step 3: Convert the volume to cubic meters

We will use the relationship 1 m³ = 10⁹ mm³.

2.81 × 10⁶ mm³ × 1 m³ / 10⁹ mm³ = 2.81 × 10⁻³ m³

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3 years ago
Electrons is an excited state are more likely to enter into chemical reactions.
Alex17521 [72]

A. True would be the best answer

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3 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

4 0
3 years ago
You need 5.0 grams of LiOH for a reaction, but all you have on hand is a 2.75 M aqueous LiOH solution. What volume of the soluti
Natalka [10]

Answer:

75.9mL of 2.75M LiOH solution

Explanation:

Molarity is an unit of concentration defined as the ratio between moles and liters. And the molar mass of LiOH is 23.95g/mol. With this information, it is possible to know the volume of solution you should add to supply 5.0g, thus:

5.0g LiOH × (1mol / 23,95g) × (1L / 2.75mol) × (1000mL / 1L) = <em>75.9mL of 2.75M LiOH solution</em>

<em />

I hope it helps!

5 0
3 years ago
What is the percentage of water in hydrated calcium chloride​
earnstyle [38]

Answer:

24.5%

Explanation:

You just add up the atomic masses.  

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This is not my answer but I found it on Yahoo answers and it was answered by Anonymous.

8 0
3 years ago
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