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Evgesh-ka [11]
2 years ago
11

As a part o a clinical study, a pharmacist is asked to prepare a modi ication o a standard 22 g package o a 2% mupirocin ointmen

t by adding the needed quantity o mupirocin powder to prepare a 3% w/w mupirocin ointment. How many milligrams o mupirocin powder are required?
Chemistry
1 answer:
son4ous [18]2 years ago
4 0

Answer:

240 mg

Explanation:

In a dilution, the total quantity of the solute remains equal, so the concentration multiplied by the volume, or mass, of it, must be constant. In this case, the concentration is measured by the mass (w/w), so:

C1M1 = C2M2

Where C is the concentration, M is the mass, 1 is the concentrated solution, and 2 the diluted. The powder mupirocin has 100% of concentration, and the diluted solution will be mass equal to 22 g plus the mass added, so:

100%*M1 = 3%*(22 + M1)

100M1 = 66 + 3M1

97M1 = 66

M1 = 0.6804 g

But the ointment already has 2% of mupirocin, which is in mass:

0.02* 22 = 0.44 g

So, the mass needed to be added is 0.6804 - 0.44 = 0.24 g = 240 mg

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Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at
asambeis [7]

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

6 0
2 years ago
You wish to make a 0.339 m h2so4 solution from a stock solution of 12.0 m h2so4. how much concentrated acid must you add to obta
Alex_Xolod [135]
When preparing diluted solutions from concentrated solutions , we can use the following equation;
c1v1 =c2v2
Where c1 and v1 are the concentration and volume of the concentrated solution
c2 is the concentration of the diluted solution to be prepared
v2 is the volume of the diluted solution
Substituting the values;
12.0 M x v1 = 0.339 M x 100 mL
v1 = 2.825 mL needs to be taken from the stock solution
8 0
2 years ago
The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the or
Aloiza [94]

Answer:

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life  Carbon 14 =  5730 year

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{5730}\ hour^{-1}

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,  

\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}

\frac {[A_t]}{[A_0]}=0.007015

<u>In percentage, the sample of C-4 remains = 0.7015 %</u>

3 0
2 years ago
How many grams are in 0.5 moles of C10H16?<br><br>Please show your work for marked brainliest!​
vagabundo [1.1K]

Answer:

68 g

Explanation:

Molar mass (C10H16) = 10*12.0 g/mol + 16*1.0 g/mol = (120+16)g/mol =

= 136 g/mol

m (C10H16) = n(C10H16)*M(C10H16) = 0.5 mol*136 g/mol = 68 g

n(C10H16) - number of moles of C10H16

M(C10H16) - molar mass of C10H16

5 0
2 years ago
How much NaAlO2 (sodium aluminate) is required to produce 5.01 kg of Na3AlF6?
Paha777 [63]
Balance Chemical equation is as follow,

                        3NaAlO₂ + 6HF   →   Na₃AlF₆ + 3H₂O + Al₂O₃

According to this equation,

      209.94 g of Na₃AlF₆ is produced by reacting = 245.91 g of NaAlO₂

So,

      5010 g of Na₃AlF₆ will be produced by reacting = X grams of NaAlO₂

Solving for X,
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                        X  =  5868 g
Or,
                        X  =  5.868 Kg of NaAlO₂
7 0
2 years ago
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