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3 years ago

The table shows the thickness, top density, and bottom

Chemistry

2 answers:

Dominik [7]3 years ago

8 0

Answer:

Answer is B on edge

Explanation:

OleMash [197]3 years ago

3 0

$\huge{\textbf{\textsf{{\color{pink}{An}}{\red{sw}}{\orange{er}} {\color{yellow}{:}}}}}$

As depth increases, the density of the layers decreases.

Thanks.
Hope it helps.

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75 mL of water is added to a 360 mL solution of acetic acid with a concentration of 0.87 M. Determine the molarity of the new so

kramer

Answer:

The molarity of the new solution is 0.72 M

Explanation:

Step 1: Data given

Volume of the original solution = 360 mL =.360 L

Molarity = 0.87 M

We add 75 mL = 0.075 L

Step 2: Calculate moles

Moles = molarity * volume

Moles = 0.87 M * 0.360 L

Moles = 0.3132 moles

Step 3: Calculate new molarity

The number of moles stays constant

Molarity = moles / volume

Molarity = 0.3132 moles / (0.36+0.075)

Molarity = 0.3132 moles / 0.435 L

Molarity = 0.72 M

The molarity of the new solution is 0.72 M

3 0

3 years ago

Is it possible to have a bent molecule without a molecular dipole?

s2008m [1.1K]

Answer:

no but that is what we were taught

3 0

3 years ago

Write short note on

WITCHER [35]

I hope it helps you ❤️❤️❤️❤️

4 0

3 years ago

Read 2 more answers

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

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

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)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

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

For propanoic acid:

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:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 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:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

Zinc, sulphur, oxygen What would this compound be called?

Nataliya [291]

Zinc sulphate

Zinc sulphateZnSO₄

This is the answer.

6 0

2 years ago