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jenyasd209 [6]
4 years ago
8

An ibuprofen suspension for infants contains 100 mg/5.0 mL suspension. The recommended dose is 10 mg/kg body weight. How many mL

of this suspension should be given to an infant weighing 18 lb?
Chemistry
1 answer:
iren [92.7K]4 years ago
5 0

Answer:

32,8mL

Explanation:

1lb=2.2kg

18lb= 8.2 kg  

100mg/5mL=20mg/1mL

164mg=32.8mL

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Irina-Kira [14]

Answer:

b

Explanation:

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3 years ago
Based on the information in the table, which of the following arranges the bonds in order of decreasing polarity?
patriot [66]

Polarity is the chemical property based on the electric charge and orientation of the poles. Al−O>H−Br>As−S is arranged in decreasing order of polarity. Thus, option d is correct.

<h3>What is polarity?</h3>

Polarity is a chemical property of the distribution of the electrical charges over their respective atom in the molecule joined by the bonds. The relation between the polarity and the difference in electronegativity is directly proportional.

The electronegativity difference between the elements are:

  • Al−O = 1.8
  • H−Br = 0.8
  • As−S = 0.4

As the electronegativity difference between Al−O = 1.8 is the highest it will have the highest polarity followed by H−Br = 0.8, and As−S = 0.4, with the lowest polarity.

Therefore, option D. Al−O>H−Br>As−S is arranged in decreasing order of polarity.

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6 0
2 years ago
How many truck stop by the store on Wednesday
siniylev [52]

Answer:

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Explanation:

7 0
2 years ago
Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr
lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

ΔT_{f} = iK_{f}m

where:

ΔT_{f} = change in freezing point (unknown)

i = Van't Hoff factor

K_{f} = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = \frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }

molal concentration = 0.1219m

Now, putting in the values to the equtaion ΔT_{f} = iK_{f}m we get,

ΔT_{f} = 2 × 1.86 × 0.1219

ΔT_{f} = 0.4536°C

So, ΔT_{f} of solution is,

ΔT_{f_{solution} } = 0.00°C - 0.45°C

ΔT_{f_{solution} } =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

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7 0
2 years ago
How much heat energy is required to raise the temperature of 15g of silver from 25°C to 55°C, assuming the specific heat of silv
Alenkasestr [34]

Answer:

108 j

Explanation:

.24 j/g-C  * 15 g * (55-25) =

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2 years ago
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