The answer to this question is 6.25ml
To answer this question, you need to calculate the azithromycin drug doses for this patient. The calculation would be: 25kg * 10mg/kg/d= 250mg/d
Then multiply the doses with the available drug. It would be:
250 mg/d / (200mg/5ml)= 6.25ml/d
Really good question i hope to see the answer to
<span>Fe(OH)3(S) +3HNO3(aq)----->Fe(NO3)3(aq) + 3H20(aq)
M(Fe(OH)3)=56+48+3=107; M(HNO3)= 48+14+1=63
n(Fe(OH)3)=5.4/107=0.05; n(HNO3)=2.6/63=0.04
n(Fe(OH)3):n(HNO3)=1:3, which means that the HNO3 should be three times (molar) than the Fe(OH)3, but you can see that it is, actually, even less than the Fe(OH)3, meaning that HNO3 is the limiting reagent and the amount of Fe(OH)3 which is going to react with HNO3 is 0.04/3=0.013 i.e. 0.05-0.013=0.037 mol Fe(OH)3 is left after the completion.
Just in case you can convert it into mass, but I suppose this is enough.</span>
Explanation:
Average atomic mass of Chlorine = 35.45 amu
The percentage of isotopes is the percentage abundance.
This is given by;
Average atomic mass = (Mass of isotope * Percentage abundance of 35/17 Cl) + (Mass of isotope * Percentage abundance 37/17 Cl)
Let percentage abundance of 35/17 Cl = x
Then percentage abundance of 37/17 Cl = 1 - x
Putting the values in the equation;
35.45 = (35 * x) + (37 * (1-x))
35.45 = 35x + 37 - 37x
35.45 - 37 = 35x - 37x
- 1.55 = -2x
x = 1.55 / 2 = 0.775
Percentage abundance of 35/17 Cl = 0.775 * 100 = 77.5%
Percentage abundance of 37/17 Cl = (1 - 0.775) * 100 = 22.5%
Answer:
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Explanation: