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Alex17521 [72]
3 years ago
9

PLEASE HELP!)NO LINKS PLEASE Rhonda created a table to classify materials that conduct electricity and materials that do not con

duct electricity. Which table correctly classifies these materials?
Materials That Conduct Electricity Materials That Do Not Conduct Electricity
steel plastic
copper glass
gold rubber
aluminum wood

Materials That Conduct Electricity Materials That Do Not Conduct Electricity
steel plastic
gold glass
aluminum rubber
wood copper

Materials That Conduct Electricity Materials That Do Not Conduct Electricity
plastic gold
rubber steel
aluminum glass
wood copper

Materials That Conduct Electricity Materials That Do Not Conduct Electricity
plastic gold
rubber steel
glass aluminum
wood copper
Chemistry
1 answer:
Bingel [31]3 years ago
7 0

Answer:

basically all but the....... second or fourth please dont report if wrong but i know most objects on the list can not be eletrically conducted im sorry oh and hi abbey!!!

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For the n = 3 electron shell, which of the following quantum numbers are valid? Check all that apply.
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Answer:

The valid quantum numbers are l=0, l=-2 and l= 2.

Explanation:

Given that,

n = 3 electron shell

Suppose, the valid quantum numbers are

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m = –2

l = –1

m = 2

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The value of n = 3

Principle quantum number :

Then the principal quantum number is 3. Which is shows the M shell.

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The azimuthal quantum number is l.

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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
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<u>Answer:</u> The pH of acid solution is 4.58

<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 KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

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Hence, the pH of acid solution is 4.58

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