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kumpel [21]
2 years ago
8

Azithromycin suspension dosing is 10 mg/kg/d and is available in 200mg/5ml doses. how many ml would a 25kg child receive daily?

7.5 ml 5.0 ml 6.25 ml 6.0 ml
Chemistry
1 answer:
bekas [8.4K]2 years ago
8 0
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
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If you compared 1 m solutions, was a 1 m nacl solution more or less hypertonic than a 1 m sucrose solution? what is your evidenc
igor_vitrenko [27]
When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:

P = iMRT,
for strong electrolytes, i = number of ions. 
for nonelectrolytes, i = 1

1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa

The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa

<em>So, that means that NaCl is more hypertonic than the sucrose solution.</em>

2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
<em>In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.</em>

7 0
2 years ago
A solution contains one or more of the following ions: Ag + , Ca 2 + , and Co 2 + . Ag+, Ca2+, and Co2+. Lithium bromide is adde
Eva8 [605]

Answer:

Since with LiBr no precipitation takes place. So, Ag+ is absent

When we add Li2SO4 to it, precipitation takes place.

Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate

Thus, Ca2+ is present.

When Li3PO4 is added, again precipitation takes place.Reaction is:

Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate

A. Ca2+ and Co2+ are present in solution

B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)

C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)

8 0
2 years ago
A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
rusak2 [61]

Answer:

Explanation:

Given parameters:

Initial temperature T₁  = 25.2°C  = 25.2 + 273  = 298.2K

Initial pressure  = P₁  = 0.6atm

Final temperature = 72.4°C   = 72.4 + 273  = 345.4K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

                  \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

          \frac{0.6}{298.2}   = \frac{P_{2} }{345.4}  

          P₂   = 0.7atm

         

3 0
2 years ago
Acetaminophen (pictured) is a popular nonaspirin, "over-the-counter" pain reliever. what is the mass % (calculate to 4 significa
77julia77 [94]

Acetaminophen as a chemical formula of C8H9NO2. The molar masses are:

C8H9NO2 = 151.163 g/mol

C = 12 g/mol

H = 1 g/mol

N = 14 g/mol

O = 16 g/mol

 

<span>TO get the mass percent, simply multiply the molar mass of each elements  with the number of the element divide by the molar mass of acetaminophen, that is:</span>

%C = [(12 * 8) / 151.163] * 100% = 63.50%

%H = [(1 * 9) / 151.163] * 100% = 5.954%

%N = [(14 * 1) / 151.163] * 100% = 9.262%

<span>%O = [(16 * 2) / 151.163] * 100% = 21.17% </span>

8 0
2 years ago
Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bee
lora16 [44]

Answer:

The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

b) 9.28 x 10 ¹⁵ atoms of O2

Explanation:

MW C7H14O2 = 84 + 14 + 32 = 130 g

a)        130 g of C7H14O2 ---------------- 1 mol of C7H14O2

           1 x 10 ⁻⁶ g              ---------------      x

           x = 7.7 x 10 ⁻⁹ mol

          1 mol of C7H14O2   --------------   6 .023 x 10 ²³ molecules

          7.7 x 10⁻⁹ mol          --------------    x

          x = 4.64 x 10¹⁵ molecules

b)      130 g of C7H14O2   ----------------   1 mol of C7H14O2

         1 x 10⁻⁶  C7H14O2   -----------------     x

         x = 7.7 x 10 ⁻⁹ mol of C7H14O2

        1 mol of C7H14O2    ---------------   2 mol of O2

        7.7 x 10 ⁻⁹                 ----------------   x

         x = 1.54 x 10⁻⁸ mol of O2

       1 mol of O2 -----------------  6.023 x 10 ²³ atoms

       1.54 x 10 ⁻⁸  ----------------   x

        x = 9.28 x 10 ¹⁵ atoms of O2

8 0
2 years ago
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