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Romashka-Z-Leto [24]
3 years ago
8

A student wants to make 50-mL of a 0.6 mg/L solution of acetic acid. She will be making the solution by diluting a stock with a

concentration of 2.0 mg/L. Put the steps for her dilution in order. 4. Cap the flask and invert several times to mix. 2. Fill the volumetric flask with deionized water until the meniscus touches the mark. 6. Swirl the flask to mix the contents. 1. Obtain a 50-mL volumetric flask. 3. Use a pipet to add the appropriate volume of stock solution to the flask. 2. Add approximately 10 mL of deionized water to the empty flask.
Chemistry
1 answer:
IrinaVladis [17]3 years ago
5 0

Answer:

The steps in order are;

1. Obtain a 50-mL volumetric flask

2. Add approximately 10 mL of deionized water to the empty flask

3. Use a pipet to add the appropriate volume of stock solution to the flask

4. Cap the flask and invert several times to mix

5. Fill the volumetric flask with deionized water until the meniscus touches the mark

6. Swirl the flask to mix the contents

The volume of the acetic acid to be added to the solution is 15 mL

Explanation:

The given steps are;

4. Cap the flask and invert several times to mix

5. Fill the volumetric flask with deionized water until the meniscus touches the mark

6. Swirl the flask to mix the contents

1. Obtain a 50-mL volumetric flask

3. Use a pipet to add the appropriate volume of stock solution to the flask

2. Add approximately 10 mL of deionized water to the empty flask

The given parameters are;

The volume of the volumetric flask = 50-mL

The concentration of the final solution of acetic acid = 0.6 mg/L

The concentration of the stock solution = 2.0 mg/L

Let 'x' represent the volume of the stock solution added, we have;

2.0 mg/L·x + (50 - x)×0 = 0.6 mg/L × 50-mL

x = 0.6 mg/L × 50-mL/(2.0 mg/L) = 15 mL

The volume of the stock solution to be added, x = 15 mL

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6

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The atomic number for phosphorous is 15, meaning that it has 15 electrons (and protons). The first and second shells would be filled up with 2 and 8 electrons respectively, leaving 5 which goes on the third shell, which is also the valence shell, meaning phosphorous has 5 valence electrons.

Since the atomic number of sulfur is 16, the first and second shells are also filled up with 2 and 8 electrons respectively, leaving 6 to be on the third shell, the valence shell. Hence, sulfur has 6 valence electrons.

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Calculate the how many sodium atoms must react completely to give 33,6 dm³ hydro-
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To give 33.6 dm³ hydrogen gas at STP, 18.06 x 10²³ atoms of Na must react completely.

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A mole is a unit of measurement used to measure the amount of any fundamental entity (atoms, molecules, ions) present in the substance.

As according to the given equation, 2 moles (ie 12.04 x 10²³ atoms) of Na-atoms produces 1 mole (22.4 ltr) of H₂-gas.

Hence, to produce 33.6 ltr (equivalent to 33.6 dm³) of H₂-gas produced by ;

   = 12.04 x 10²³ atoms of Na / 22.4 ltr of H₂-gas x 33.6 ltr

   = 18.06 x 10²³ atoms of Na

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How does electronegativity impact polarity of molecules?
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What is the answers and pls show work if possible!!
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D = m / V


It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...


V = L x W x H

Volume = Length x Width x Height


start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g


V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3


Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

1 ft = 30.48 cm


Compute the height (thickness)

V = LxWxH

H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm

H = 4.64 x 10^-6 cm


Convert to nanometers

4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm


Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.


Atomic radius gold = 174 pm

Diameter = 348 pm


46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold


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