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3 years ago

How many mL of a 4.50 (m/v)% solution would contain 23.0 g of solute?

Chemistry

2 answers:

nlexa [21]3 years ago

4 0

23.0 g/ (4.50 g/ mL) gives 5.11 mL as the volume.

Helen [10]3 years ago

3 0

<u>Answer:</u> The given amount of solute will be present in 511.11 mL of solution.

<u>Explanation:</u>

We are given:

Concentration of solution = 4.50 (m/v) %

This means that 4.50 grams of solute is present is 100 mL of solution.

We need to find the volume of solution that can contain 23.0 grams of solute.

By applying unitary method, we get:

4.50 grams of solute is present in 100 mL of solution.

So, 23.0 grams of solute will be present in = $\frac{100mL}{4.50g}\times 23.0=511.11mL$ of solution.

Hence, the given amount of solute will be present in 511.11 mL of solution.

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How many molecules are in 41.8 g of sulfuric acid

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Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

$Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol$

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

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Therefore, 0.426187053 moles of sulfuric acid is equal

$\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules$

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

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1 year ago

Acids are described as corrosive because they a. turn litmus blue. b. taste bitter. c. “eat away” at other materials. d. feel sl

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3 years ago

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2 years ago

What is the ph of a 0.25 m solution of c6h5nh2 given that its kb is 1.8 x 10-6?

IrinaK [193]

The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.

<h3>How do we calculate pH of weak base?</h3>

pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:

pH = pKb + log([HB⁺]/[B])

pKb = -log(1.8×10⁻⁶) = 5.7

Chemical reaction for C₆H₅NH₂ is:

C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻

Initial: 0.25 0 0

Change: -x x x

Equilibrium: 0.25-x x x

Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]

Kb = x² / 0.25 - x

x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:

1.8×10⁻⁶ = x² / 0.25

x² = (1.8×10⁻⁶)(0.25)

x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]

On putting all these values on the above equation of pH, we get

pH = 5.7 + log(0.67×10⁻³/0.25)

pH = 3.13

Hence pH of the solution is 3.13.

To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361

#SPJ4

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2 years ago