Answer:
The answer to your question is 3 ml
Explanation:
Data
Dosage = 9.0 mg/ kg
Child's weight = 42.9 pounds
Suspension = 60 mg/ml
milliliters = ?
Process
1.- Convert the weight to kg
1 pound ------------------- 0.453 kg
42.9 pounds --------------- x
x = (42.9 x 0.453) / 1
x = 19.43 kg
2.- Calculate the milligrams the child needs
1 kg of weight ------------ 9 mg
19.43 kg ---------------------- x
x = (19.43 x 9) / 1
x = 174.87 mg of oxcarbazepine
3.- Calculate the milliliters needed
60 mg of suspension ------------- 1 milliliters
174.87 mg -------------- x
x = (174.87 x 1) / 60
x = 2.9 ml ≈ 3 ml
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How does one determine the identity and structure of an unknown compound? This is not a trivial task. Modern x-ray and spectroscopic techniques have made the job much easier, but for some very complex molecules, identification and structure determination remains a challenge. In addition to spectroscopic information and information obtained from other instrumental methods, chemical reactions can provide useful structural information, and physical properties can contribute significantly to confirming the identity of a compound.
In this experiment, you will be asked to identify an unknown liquid, which will be either an alcohol, aldehyde, or ketone. Identification will be accomplished by carrying out chemical tests, called classification tests, preparing a solid derivative of the unknown and determining its melting point (MP), making careful observations, and analyzing the NMR spectrum of the unknown.
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Answer:
The log will float on the water because its density is lower than the liquid, so it will stay at the top due to Archimedes' principle.
Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
![0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)](https://tex.z-dn.net/?f=0.80%3D%5Cfrac%7B1%7D%7B95%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.86%29%7D%5Cright%29)
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M
