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Gekata [30.6K]
3 years ago
15

As a part of a clinical study, a pharmacist is asked to prepare a modification of a standard 22g package of a 2% mupirocin ointm

ent by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment. How many mg of mupirocin powder are required?
Chemistry
1 answer:
Vesna [10]3 years ago
8 0

Answer:

226.8 mg of mupirocin powder are required

Explanation:

Given that;

weight of standard pack = 22 g

mupirocin by weight = 2%

so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g

so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment

mg of mupirocin powder are required = ?, lets rep this with x

Total weight of ointment = 22 + x g

Amount of mupirocin  = 0.44 + x g

percentage of mupirocin  in ointment is 3?

so

3/100 = 0.44 + x g / 22 + x g

3( 22 + x g ) = 100( 0.44 + x g )

66 + 3x g = 44 + 100x g

66 - 44 = 100x g - 3x g

97 x g = 22

x g = 22 / 97

x g = 0.2268 g

we know that; 1 gram = 1000 Milligram

so 0.2268 g = x mg

x mg = 0.2268 × 1000

x mg  = 226.8 mg

Therefore, 226.8 mg of mupirocin powder are required

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Answer:

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Answer:

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Explanation:

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There are two Al three zinc and six chlorine atoms on both side of equation so it is correctly balanced.

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This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

7 0
3 years ago
If 0.450 moles of iron III oxide (Fe2O3) are allowed to react with an excess of aluminum (Al) and 43.6 grams of iron (Fe) is pro
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