A. Quantity of saline = 500mL
Rate of infusion = 80 mL / h
Infusion time = Quantity / Rate = 500 mL / (80 mL/hr) = 6.25 hr
b. Child weight = 72.6 lb = 32.93 kg
Medrol to be given = 1.5 mg per kg
Quantity of Medrol = 20 mg/mL
Dosage available = 20 mg/mL / 1.5 mg/kg = 13.33 kg/mL
Dosage according to body weight = 32.93 kg / 13.33 kg/mL = 2.47 mL
Answer: 
Explanation:
Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant.
All zero’s between integers are always significant.
All zero’s after the decimal point are always significant.
All zero’s preceding the first integers are never significant.
Thus has three significant figures
Answer:
No, the puddle was formed because of the sun, because if there was snow and it rained then it would have turned slippery or icy
Answer:
The pH of the buffer is 7.0 and this pH is not useful to pH 7.0
Explanation:
The pH of a buffer is obtained by using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer</em>
<em>The pKa of acetic acid is 4.74.</em>
<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>
<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>
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Replacing:
pH = 4.74 + log [0.1779mol] / [0.001mol]
<em>pH = 6.99 ≈ 7.0</em>
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The pH of the buffer is 7.0
But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,
this pH is not useful to pH 7.0
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