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4 years ago

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Anesthetic gases. one type of gas mixture used in anesthesiology

1 answer:

netineya [11]4 years ago

6 0

I dont know i don’t understand

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Sort the following analytical techniques as either classical methods or instrumental methods.

Aleonysh [2.5K]

Answer:

Instrumental methods

surface analysis

high performance liquid chromatography

atomic spectroscopy

potentiometry

Classical methods

precipitation titration

gravimetric analysis

Explanation:

Instrumental methods of analysis are those analytical methods in which the responsibility of detection has been removed from human beings and placed on automated instruments while classical methods are those analytical methods in which the responsibility of detection remains the responsibility of human beings.

Many instrumental methods such as HPLC rely on computer screens as readout devices.

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3 years ago

An unknown solution has a ph of 2. How would you classify this solution

sammy [17]

It’s an acid and it’s conducive

4 0

3 years ago

If you have 1 mole of Hydrogen, how much Hydrogen gas do you have in liters?

natita [175]

Answer:

22.4 liters

Explanation:

is this answer

7 0

3 years ago

Earth's lithospheric plates

mr_godi [17]

The answer is - D. are constantly in motion

These plates move back and forth and sometimes even end up going one above other

6 0

4 years ago

What is the pH of a buffer solution upon mixing 15.0 mL of 0.40 M HCl and 20.0 mL of 0.50 M NH? Kb (NH3) = 1.8 x 10 E. 7.00 A. 9

vladimir2022 [97]

<u>Answer:</u> The pH of resulting solution is 9.08

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ........(1)

<u>For HCl:</u>

Molarity of HCl = 0.40 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

$0.40M=\frac{\text{Moles of HCl}\times 1000}{15.0mL}\\\\\text{Moles of HCl}=0.006mol$

<u>For ammonia:</u>

Molarity of ammonia = 0.50 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

$0.50M=\frac{\text{Moles of ammonia}\times 1000}{20.0mL}\\\\\text{Moles of ammonia}=0.01mol$

The chemical reaction for hydrochloric acid and ammonia follows the equation:

$HCl+NH_3\rightarrow NH_4Cl$

Initial: 0.006 0.01

Final: - 0.004 0.006

Volume of solution = 15.0 + 20.0 = 35.0 mL = 0.035 L (Conversion factor: 1 L = 1000 mL)

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_3]})$

We are given:

$pK_b$ = negative logarithm of base dissociation constant of ammonia = $-\log (1.8\times 10^{-5})=4.74$

$[NH_4Cl]=\frac{0.006}{0.035}$

$[NH_3]=\frac{0.004}{0.035}$

pOH = ?

Putting values in above equation, we get:

$pOH=4.74+\log(\frac{0.006/0.035}{0.004/0.035})\\\\pOH=4.92$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4.92=9.08$

Hence, the pH of the solution is 9.08

3 0

4 years ago