Answer:
1.041g is the maximum sigle dose for the person
Explanation:
The maximum dose for acetaminophen is 15.0mg / kg of body weight for adults
To find the maximum single dosage for a person who weighs 153lb we must convert the lb to kg (1lb = 0.4536kg):
153lb * (0.4536kg / 1lb) =<em> 69.4kg is the mass of the person</em>
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As the maximum dose is 15.0mg / kg, the dose of the person is:
69.4kg * (15.0mg acetaminophen / kg) =
<h3>1041mg = 1.041g is the maximum sigle dose for the person</h3>
Answer 0.5
when reporting calculations from sig figs, report the number of sig figs in the least accurate measurement
so
((2.0265-2.02)/2.0265) x 100.00
((2.0265/2.0265) - (2.02/2.03)) X 100.00
(1.0000 -0.995) X 100.00
0.005 X 100.00
0.5
Answer:
pH = 4.71
Explanation:
We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:
pH = pKa + log [CH3COONa] / [CH3COOH]
<em>Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.</em>
As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:
pH = pKa + log 1
4.74 = pKa
To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl
Moles CH3COONa: Initial CH3COONa - Moles HCl.
<em>Moles CH3COOH: </em>
0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =
0.052 moles CH3COOH
<em>Moles CH3COONa: </em>
0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =
0.048 moles CH3COONa
Using H-H equation:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
<h3>pH = 4.71</h3>
