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
Hey there !
The answer is C. build up of electrical charges by friction
Static electricity is a stationary electric charge, typically produced by friction, which causes sparks or crackling or the attraction of dust or hair.
Hope this helps !
Answer:
True
Explanation:
Here is an example: chemical properties include flammability, toxicity, acidity, reactivity. we observe the changes of these properties. Therefore, It's true.
Answer:
There are approximately 1.54 moles in a 275 g sample of
.
Explanation:
To find out number of moles, fistly we have to calculate molecular mass of
.
There are 2 atoms of Potassium 1 atom of Chromium and 3 atoms of oxygen in the given compound.
For molecular mass we have to add the value of mass of 2 atoms of Potassium with mass of 1 atom of Chromium and with mass of 3 atoms of oxygen.
Atomic mass of Potassium = 39
Atomic mass of Chromium = 52
Atomic mass of Oxygen = 16
Now,
Molecular mass of
= 
The molecular mass of a compound is the mass of compound in one mole.
To find out the number of moles, we have to divide given mass of compound by its molecular mass.


Hence the number of moles in 275 gm of
is 1.54.