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adell [148]
3 years ago
9

A 38-lb child is prescribed acyclovir for chicken pox in an amount of 80 mg/ kg body weight per day to be divided in four doses

each tablet contains 700 mg of the medicine. How many tablets should be given per day? Per dose
Chemistry
1 answer:
Svetach [21]3 years ago
5 0

Answer:

\large \boxed{\text{Two tablets per day; half a tablet per dose}}

Explanation:

1. Convert pounds to kilograms

\text{Body weight} = \text{38 lb} \times \dfrac{\text{1 kg}}{\text{2.205 kg}} = \text{17.2 kg}

2. Convert kilograms of body weight to milligrams of acyclovir

\text{Mass of acyclovir} = \text{17.2 kg} \times \dfrac{\text{80 mg acyclovir}}{\text{1 kg}} = \text{1380 mg acyclovir }

3. Convert milligrams of acyclovir to number of tablets

\text{No. of tablets} = \text{1380 mg} \times \dfrac{\text{1 tablet}}{\text{700 mg}} = \textbf{2 tablets}

4. Calculate the tablets per dose

The child must receive two tablets in four doses.

\text{Tablets per dose} = \dfrac{\text{2 tablets}}{\text{4 doses}} = \textbf{0.5 tablets per dose}\\\\\text{The child must receive $\large \boxed{\textbf{two tablets per day and half a tablet per dose}}$}

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Explanation:

In general, adding H⁺ ions to any aqueous solution ALWAYS causes pH values to fall ( decrease ). Just as adding OH⁻ ions to an aqueous solution causes pH values to rise ( increase ).

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Just to support the above statement about adding OH⁻ ions showing an increase in pH values, the following is also provided FYI ..

Given 0.01M NaOH(aq) => 0.01M OH⁻(aq) + Na⁺(aq) => pOH = -log(0.01) = 2.00 => pH = 14 - pOH = 14 - 2 = 12

Increase [OH⁻] by 10x => 0.10M OH⁻(aq) => pOH = -log[OH⁻] = -log(0.10) = 1.00 => pH = 14 - pOH = 14 - 1 = 13

Increasing [OH⁻] by 10x => <u>increasing pH by 1 unit. </u>

Solution with higher H⁺ concentration shows pH decreasing by 1 unit.

______________________________________________________

Remember, for <u>any</u> aqueous solution ...

=> Adding H⁺   => always decreases pH

=> Adding OH⁻ => always increases pH

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