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shutvik [7]
3 years ago
11

A pharmacist wishes to mix a solution that is 5% Minoxidil. She has on hand 80 ml of a 4% solution and wishes to add some 7% sol

ution to obtain the desired 5% solution. How much 7% solution should she add? ...?
Chemistry
2 answers:
Alona [7]3 years ago
6 0
<span>The 80 ml solution has (.04)(80) ml of the medicine, or 3.3 ml of medicine.The pharmacist adds V ml of 7% solution, or .07V ml of medicine. The total amoune of medicine is (.07V + 3.2) mg. The total volume of solution is (V + 80) ml, so (.07V + 3.2) / (V + 80) = .05. Solve for V.</span>
Debora [2.8K]3 years ago
6 0

The 80 ml solution has (.04)(80) ml of the medicine, or 3.3 ml of medicine.The pharmacist adds V ml of 7% solution, or .07V ml of medicine. The total amoune of medicine is (.07V + 3.2) mg. The total volume of solution is (V + 80) ml, so (.07V + 3.2) / (V + 80) = .05.

Solve for V.

<span>V = 40 ml.</span>

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7 0
2 years ago
What is my theoretical yield (in moles) of Potassium Bromide (KBr) if I start with 40 grams of Iron (II) Bromide [FeBr2]? moles
Verizon [17]
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3 years ago
How does this process appear in the atmosphere?
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8 0
3 years ago
Read 2 more answers
If the half-life of a radioactive element is 4 days, how long will it take for three- fourths of a sample of the element to deca
Jlenok [28]

Answer:

\boxed{\text{8 da}}

Explanation:

The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"

The half-life of the element is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:

\begin{array}{cccl}\textbf{No. of} & & \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/da} & \textbf{remaining} & \\1 & 4 & \dfrac{1}{2} & \\\\2 & 8 & \dfrac{1}{4}& \\\\3 & 12 & \dfrac{1}{8}& \\\end{array}

\text{We see that 8 da is two half-lives, and the fraction of the element remaining is $\frac{1}{4}$.}\\\text{It takes $\boxed{\textbf{8 da}}$ for three-fourths of the element to decay}

3 0
3 years ago
Balancing nuclear equations is slightly different than balancing chemical equations. The major difference is that in nuclear rea
vesna_86 [32]

Answer:

The correct answer is - 4.

Explanation:

As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -

The sum of the product mass number of products = mass of reactant

237Np93 →233 Pa91 +AZX is the equation,

Solution:

Mass of reactants = 237

Mass of products are - Pa =233 and A = ?

233 + A = 237

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A = 4

So the equation will be:

237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)

5 0
3 years ago
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