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Afina-wow [57]
1 year ago
9

Dosage calculation order: 3 mg available: 2 mg per 6 ml how many ml will be given?

Chemistry
1 answer:
Anna35 [415]1 year ago
3 0

9ml will be given for the case of dosage calculation order: 3 mg available: 2 mg per 6 ml

Conversion factors are necessary for dosage calculation, such as when translating from pounds to kilograms or liters to milliliters. This approach, which is straightforward in design, enables physicians to deal with different units of measurement and convert factors to arrive at the solution.

dosage calculation techniques serve as a second or third check on the accuracy of the previous computation techniques. Dimensional Analysis, Ratio Proportion, and Formula or Desired Over Have Method are the three main approaches for dosage calculation. dosage calculations are frequently prescribed and labeled based on their weight or, for solutions, their strength, which is the amount of weight dissolved or suspended in a given volume.

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The ionic charge of an element in Group 6A is 2–.<br><br> -True<br><br> -False
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True. These ions are of a 2- charge. Oxygen is an example. It will form a 2- charge if ionize.  

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How many moles are in 2.5L of 1.75 M Na2CO3
Natasha_Volkova [10]

Answer:

4.4 mol.

Explanation:

Hello!

In this case, since the formula for calculating the molarity is:

M=n/V

Whereas n stands for moles and V for the volume in liters; we can solve for n as shown below when we are given the volume and the molarity:

n=V*M

Thus, we plug in the given data to obtain:

n=2.5L*1.75mol/L=4.4mol

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4 0
2 years ago
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A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc
Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

<em>2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]</em>

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

<h3>pH = 2.83</h3>

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2 years ago
The phase of matter can be changed by:
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Answer: A

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