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Digiron [165]
1 year ago
9

How many grams o an ointment base must be added to 45 g o clobetasol (T EMOVAT E) ointment, 0.05% w/w, to change its strength to

0.03% w/w?
Chemistry
1 answer:
saw5 [17]1 year ago
3 0

Answer:

Drug calculation

If we have 45g of clobetasol  = 0.05%w/w

Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g

It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w

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What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?
tankabanditka [31]
The molarity is the number of moles in 1 L of the solution. 
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol 
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
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A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the s
yKpoI14uk [10]

Answer:

27.0

Explanation:

Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.

7 0
2 years ago
Which milligram quantity contains a total of four significant figures
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D has a total of four significant figures.
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The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera
marshall27 [118]
Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2  cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius
8 0
1 year ago
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

7 0
2 years ago
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