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nikklg [1K]
2 years ago
13

Compare amplitudes, wavelengths, and frequencies of waves

Chemistry
1 answer:
mr_godi [17]2 years ago
8 0

Answer:

The answer is option b.

Explanation:

Amplitude is the distance apart each wave is.

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What is true about two neutral atoms of the element gold
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I'm pretty sure it's they have the same properties, because they don't have different masses, they don't both have missing nuclei, and they don't hav a different # of protons. Can't recall much else about the two, but I distinctly remember that
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Features passed from parent to offspring, such as eye color, is an example of what?
Olenka [21]
Parents pass characteristics such as hair color, nose shape, and skin color to their offspring. Not all of the parents' characteristics will appear in the offspring, but the characteristics that are more likely to appear can be predicted.
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3 years ago
A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction
Molodets [167]

<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

<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}}{\text{Volume of solution (in L)}}     .....(1)

<u>For NaOH:</u>

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L   (Conversion factor:   1 L = 1000 mL)

Putting values in equation 1, we get:

0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol

<u>For sulfuric acid:</u>

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol

The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 2.925\times 10^{-3} moles of KOH will react with = \frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = (4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, 2.925\times 10^{-3} moles of KOH will produce = \frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol of sodium sulfate

  • <u>For sodium sulfate:</u>

Moles of sodium sulfate = 1.462\times 10^{-3}moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M

  • <u>For sulfuric acid:</u>

Moles of excess sulfuric acid = 3.498\times 10^{-3}mol

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M

  • <u>For NaOH:</u>

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of NaOH}=\frac{0}{0.050}=0M

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

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3 years ago
Describe a method of Municipal treatment of water with neat diagram?​
stich3 [128]
Water treatment process
4 0
2 years ago
A/1 * B/C =D
Novay_Z [31]
<h3>Answer:</h3>

                  0.64 Moles of Propane

<h3>Explanation:</h3>

Data:

         Moles of Carbon  =  1.5 mol

         Conversion factor  =  7 mol C produces = 3 mol of Propane

Solution:

             As we know,

                7 moles of Carbon produces =  3 moles of Propane

Then,

            1.5 moles of Carbon will produce  =  X moles of Propane

Solving for X,

                     X =  (1.5 moles × 3 moles) ÷ 7 moles

                     X  =  0.6428571 moles of Propane

Or rounded to two significant figures,

                     X =  0.64 Moles of Propane

5 0
2 years ago
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