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pogonyaev
3 years ago
13

A physician has ordered 0.50 mg of atropine, intramuscularly. If atropine were available as 0.25 mg/mL of solution, how many mil

liliters would you need to give?
Chemistry
1 answer:
sladkih [1.3K]3 years ago
6 0
If 0.25mg of atropine is in 1mL
                  so
    0.50mg of atropine is in x

x=\frac{0.5mg*1mL}{0.25mg}=2mL
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3 years ago
A human lung at maximum capacity has a volume of 3.0 liters. If the partial pressure of oxygen in the air is 21.1 kilopascals an
MrRa [10]

Answer : 0.026 moles of oxygen are in the lung

Explanation :

We can solve the given question using ideal gas law.

The equation is given below.

PV = nRT

We have been given P = 21.1 kPa

Let us convert pressure from kPa to atm unit.

The conversion factor used here is 1 atm = 101.3 kPa.

21.1 kPa \times \frac{1atm}{101.3kPa}= 0.208 atm

V = 3.0 L

T = 295 K

R = 0.0821 L-atm/mol K

Let us rearrange the equation to solve for n.

n = \frac{PV}{RT}

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n = 0.026 mol

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3 0
3 years ago
An aqueous solution of methylamine (ch3nh2) has a ph of 10.68. how many grams of methylamine are there in 100.0 ml of the soluti
ruslelena [56]

Answer:

3.4 mg of methylamine

Explanation:

To do this, we need to write the overall reaction of the methylamine in solution. This is because all aqueous solution has a pH, and this means that the solutions can be dissociated into it's respective ions. For the case of the methylamine:

CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻     Kb = 3.7x10⁻⁴

Now, we want to know how many grams of methylamine we have in 100 mL of this solution. This is actually pretty easy to solve, we just need to write an ICE chart, and from there, calculate the initial concentration of the methylamine. Then, we can calculate the moles and finally the mass.

First, let's write the ICE chart.

       CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻     Kb = 3.7x10⁻⁴

i)            x                                      0            0

e)          x - y                                  y            y

Now, let's write the expression for the Kb:

Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂]

We can get the concentrations of the products, because we already know the value of the pH. from there, we calculate the value of pOH and then, the OH⁻:

The pOH:

pOH = 14 - pH

pOH = 14 - 10.68 =  3.32

The [OH⁻]:

[OH⁻] = 10^(-pOH)

[OH⁻] = 10^(-3.32) = 4.79x10⁻⁴ M

With this concentration, we replace it in the expression of Kb, and then, solve for the concentration of methylamine:

3.7*10⁻⁴ = (4.79*10⁻⁴)² / x - 4.79*10⁻⁴

3.7*10⁻⁴(x - 4.79*10⁻⁴) = 2.29*10⁻⁷

3.7*10⁻⁴x - 1.77*10⁻⁷ = 2.29*10⁻⁷

x = 2.29*10⁻⁷ + 1.77*10⁻⁷ / 3.7*10⁻⁴

x = [CH₃NH₂] = 1.097*10⁻³ M

With this concentration, we calculate the moles in 100 mL:

n = 1.097x10⁻³ * 0.100 = 1.097x10⁻⁴ moles

Finally to get the mass, we need to molar mass of methylamine which is 31.05 g/mol so the mass:

m = 1.097x10⁻⁴ * 31.05

<h2>m = 0.0034 g or 3.4 mg of Methylamine</h2>
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