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Vinvika [58]
2 years ago
9

The average dosage of oxcarbazepine for an epileptic child between the ages of 4 and 16 is 9.00 mg per 1 kg of body weight (9.00

mg/kg). Calculate how many milliliters of oxcarbazepine a child who weighs 37.3 pounds should be prescribed considering the medication is served as a suspension of 60.0 mg/mL. Keep in mind that 1 pound is equivalent to 0.453 kilograms.
Chemistry
2 answers:
melamori03 [73]2 years ago
8 0
Convert the child weight (37.3 pounds) to kilograms

37.3 lb x 0.453 kg /1lb = "A kg"

multiply the dose (9.00mg/kg) by the weight of the child to find how much you need to give him

A kg * 9.00 mg/1kg = "B mg" 

calculate the mL of suspension dividing the "B mg" by the concentration of the suspension 60.0 mg/mL

B mg * 1mL/ 60.0 mg = C mL <span>oxcarbazepine</span>
vovikov84 [41]2 years ago
7 0

Answer:

2.53 milliliters of oxcarbazepine medication should be given to the child.

Explanation:

The average dosage of oxcarbazepine for an epileptic child = 9.00 mg/kg

Weight of the child = 37.3 pounds = 16.9 kg

(1 pound = 0.453 kilograms)

Dosage of oxcarbazepine for 16.9 kg child:

= 9.00 mg/kg\times 16.9 kg=152 mg

Prescribed medication is served as 60.0 mg/mL which means that 60.0 mg of dose is present in 1 mL of water solution.

Then 152 mg of dose will be present in:

\frac{1 mL}{60.0}\times 152=2.53 mL

2.53 milliliters of oxcarbazepine medication should be given to the child.

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LuckyWell [14K]
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7 0
2 years ago
If the patient has to be administered a dosage of 2 tablets every 8 hours for 7 days, what is the number of tablets required for
Lilit [14]
If the patient has to take 2 tablets every 8 hours for 7 days.
 24/8=3     3*2=6
this means that he patient will have to take 6 tablets every day. 
6*7=42   And the patient must take 42 tablets in all 7 days
Hope this helps! :)
6 0
2 years ago
The vapor pressure of benzene, c6h6, is 100.0 torr at 26.1 °c. Assuming raoult's law is obeyed, how many moles of a nonvolatile
frutty [35]

According to Raoult's law:

p_{solution} = X_{solvent}\times P_{solvent}    -(1)

where p_{solution} is observed vapor pressure of the solution, X_{solvent} is mole fraction of solvent, and P_{solvent} is vapor pressure of the pure solvent.

Vapor pressure of benzene = 100 torr at 26.1^{o} C   (given)

Reduction in vapor pressure on addition of non-volatile solute = 10.0 %   (given)

So, vapor pressure of the component in the solution = 100 - 10 = 90 torr

Substituting the values in formula (1):

90 = mole fraction of benzene \times 100

mole fraction of benzene = \frac{90}{100} = 0.9

Mole of benzene = Volume\times \frac{density}{Molar Mass}

Mole of benzene = 100 cm^{3}\times \frac{0.8765 g/cm^{3}}{78 g/mol} = 1.124 mol

Since, mole fraction of benzene =\frac{mole of benzene }{mole of benzene +mole of non volatile solute }

So, 0.9 =\frac{1.124 }{1.124+mole of non volatile solute }

\frac{1.124}{0.9} =1.124+mole of non volatile solute

mole of non volatile solute = 1.249 - 1.124 = 0.125

Hence, moles of a nonvolatile solute to be added is 0.125.

3 0
2 years ago
6. A rectangular block of wood has length 6.0 cm, width 5.0 cm and height 10.0 cm. It has mass l50g. What is the mass of a block
vovikov84 [41]
The mass would also be 150g like the first block of wood. Both of their volumes are 300 cm^2, and since they are the same type of wood they have the same density. Therefore following d=m/v, they have the same mass
4 0
2 years ago
4p + 5O2 -&gt; P4O10 ; the percent yield of PO4O10 when 6.20 g of phosphorus burns into excess oxygen is 67.0%. What is the yiel
denis23 [38]

Answer:    36.9 g

Explanation:

P4 + 5O2 = P4O10  Balanced equation

moles P4 present = 23.9 g x 1 mole/123.88 g = 0.193 moles

moles O2 present = 20.8 g x 1 mol/32 g = 0.65 moles O2

From balanced equation, mole ratio O2 : P4 is 5:1.  Is 0.65 moles O2 5x 0.193 moles?  NO.  You don't have enough O2.

O2 is limiting in this reaction.

 

theoretical moles of P4O10 = 0.65 moles O2 x 1 mole P4O10/5 moles O2 = 0.13 moles P4O10

mass of P4O10 produced = 0.13 moles x 283.9 g = 36.9 g

3 0
1 year ago
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