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4 years ago

What are dilute and concentrated acids?

Chemistry

1 answer:

mars1129 [50]4 years ago

4 0

Answer:

A dilute acid is when the concentration of water mixed in the acid is higher than the concentration of the acid itself. For example, 5% sulfuric acid is a dilute acid.

A concentrated acid is an acid that is in either pure form or has a high concentration.

Hope this helps! :)

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Select the physical properties. Select all that apply.

Darya [45]

I mean the answer is -2 m/s^2

6 0

4 years ago

Read 2 more answers

An analytical chemist is titrating 242.5mL of a 1.200M solution of hydrazoic acid HN3 with a 0.3400M solution of NaOH . The pKa

fgiga [73]

Answer: The pH of the solution is 12.61

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 hydrazoic acid:

Molarity of hydrazoic acid solution = 1.200 M

Volume of solution = 242.5 mL

Putting values in equation 1, we get:

$1.200M=\frac{\text{Moles of hydrazoic acid}\times 1000}{242.5mL}\\\\\text{Moles of hydrazoic acid}=0.291mol$

For NaOH:

Molarity of NaOH solution = 0.3400 M

Volume of solution = 1006 mL

Putting values in equation 1, we get:

$0.3400M=\frac{\text{Moles of NaOH}\times 1000}{1006mL}\\\\\text{Moles of NaOH}=0.342mol$

The chemical reaction for hydrazoic acid and NaOH follows the equation:

$HN_3+NaOH\rightarrow NaN_3+H_2O$

Initial: 0.291 0.342

Final: 0 0.051 0.291 0.291

Volume of solution = 242.5 + 1006 = 1248.5 mL = 1.2485 L (Conversion factor: 1 L = 1000 mL)

For NaOH left:

Left moles of NaOH = 0.051 moles

Volume of the solution = 1.2485 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0.051mol}{1.2485L}=0.0408M$

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions

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

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.0408M$

Putting values in above equation, we get:

$pOH=-\log(0.0408)\\\\pOH=1.39$

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

$pH+pOH=14\\\\pH=14-1.39=12.61$

Hence, the pH of the solution is 12.61

8 0

3 years ago

Please which one is it, it’s due in 1 hour please help

Brilliant_brown [7]

Answer:

Sulfur

Explanation:

7 0

3 years ago

What is the correct formula for the product represented by the letter x

ANTONII [103]

Then I would guess the answer is D because Na2OH is much more common and stable than Na2O is

3 0

3 years ago

The solubility product for Zn(OH)2 is 3.0×10−16. The formation constant for the hydroxo complex, Zn(OH)2−4, is 4.6×1017. What is

Yuri [45]

Answer:

$[OH^-]= 1.12 \times 10^{-2} \ M$

Explanation:

$\text{From the information given:}$

$K_{sp} \ of \ Zn(OH)_2 \ is \ 3.0 \times 10^{-16}$

$K_f \ of \ Zn(OH)_2 = 4.6 \times 10^{17}$

$\mathtt{K_{sp} \ \ \ \ \ \ \ \ \mathtt{Zn(OH)_{2(s)}} \to Zn^{2+}_{(aq)} + 2OH^{-}_{(aq)} }$

$\mathtt{K_{f} \ \ \ \ \ \ \ \ \mathtt{Zn^{2+}_{(aq)} } + 4OH^-_{(aq)} \to Zn(OH)^{2-}_{4}_{(aq)}}$

$\mathtt{K \ \ \ \ \ \ \ \ \mathtt{{Zn(OH)_{2}_{(s)}}} + 2OH^{-}_{(aq)} \to Zn(OH)^{2-} _{4(aq) }}$

$\text{To calculate the minimum concentration of OH}^{-}$

$K = \dfrac{Zn(OH)^-_4}{[OH^-]^2} \\ \\$

$[OH^-]^2= \dfrac {0.017}{1.380 \times 10^2}$

$[OH^-]^2=1.232 \times 10^{-4}$

$[OH^-]= 1.12 \times 10^{-2} \ M$

8 0

3 years ago