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>
There are <span>4 1/2 cups of flour in a pound of flour. </span>
Answer: The pH of the solution is 11.2
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in ml
moles of 
= 
(1g=1000mg)
Now put all the given values in the formula of molality, we get
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
According to stoichiometry,
1 mole of gives 2 mole of 
Thus 0.0298 moles of gives =
moles of
Putting in the values:
![pOH=-\log[0.0596]=2.82](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B0.0596%5D%3D2.82)
Thus the pH of the solution is 11.2
Answer: The correct answer is: [B]:
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" organic acid and amines " .
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<u>Note</u>: Choice B: "organic acid and amines" ;
is the only answer choice that contains "amines" (hint: <u> amin</u><u>o acid</u> / <u>amin</u><u>e)</u> ; which are "proteins" .
As such; Choice "B" is the <u><em>only</em></u> correct answer choice.
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Hope this helps!
Best wishes to you!
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The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
