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2 years ago

How many mL of alcohol are in 240 mL of 95.0% alcohol solution?

2 answers:

5 0

To work it out, you divide 240 by 100 to work out 1% of it, then multiply that by 95 to work out 95% of it. So
(240/100) * 95 = 228mL

Marat540 [252]2 years ago

3 0

Answer:228 mL of alcohol are in 240 mL of 95.0% alcohol solution.

Explanation:

Total volume of the solution = 240 mL

Percentage of alcohol in solution = 95.0%

Volume of volume of alcohol in solution = V

$95.0\%=\frac{V}{240 mL}\times 100$

V = 228 mL

228 mL of alcohol are in 240 mL of 95.0% alcohol solution.

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Which of the following is a change from a more condensed to a less condensed state of matter?

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You are given 230 grams of a substance with a half-life of 0.75 years. How much will remain after 3 years?

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After 3 years, the substance will go through 4 half lives. You can do 230*(1/2)^4, which is 14.375.

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2 years ago

Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?.

Julli [10]

HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.

<h3>What are buffer solutions and how do they differ?</h3>

The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.

<h3>Describe buffer solution via an example.</h3>

When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
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learn more about buffer solutions here

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8 0

8 months ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

<u>Answer:</u> The theoretical yield of acetanilide is 6.5 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For aniline:</u>

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

<u>For acetic anhydride:</u>

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

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

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

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

2 years ago