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Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54-g sample of vinegar was neutralized by 30.10 mL of 0.1

tekilochka [14]

Answer:

3.26 % of vinegar is acetic acid

Explanation:

Step 1: Data given

Mass of the sample = 5.54 grams

Volume of NaOH = 30.10 mL

Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

Step 6: Calculate percent by weight of acetic acid

Mass % = ( 0.1808 / 5.54 ) *100%

Mass % = 3.26 %

3.26 % of vinegar is acetic acid

7 0

3 years ago

Hello guys can you help me on this.

RideAnS [48]

Answer:

1.Sulfur dioxide and nitrogen oxide

2.a)it forms carbonic acid

b)

3.the community can use renewable energy like solar and wind power cause they produce less pollution

5 0

3 years ago

Identify the products in the equation below (the “I” is Iodine—to differentiate capital “I” from lowercase “l”)

Inessa [10]

Answer : The products are Silver sulfide, $(Ag_2S)$ and Sodium iodide, $(NaI)$ .

Explanation :

The given balanced chemical reaction is,

$2AgI+Na_2S\rightarrow Ag_2S+2NaI$

From the given balanced reaction, we conclude that the 2 moles of silver iodide react with the 1 mole of sodium sulfide to give product as 1 mole of silver sulfide and 2 moles of sodium iodide.

In a chemical reaction, reactants are represent on the left side of the right-arrow and products are represent on the right side of the right-arrow.

Therefore, in a chemical reaction the products are Silver sulfide and Sodium iodide.

5 0

3 years ago

Ivan wants to test how the amount of water consumed each day affects the number of facial blemishes people receive over a month.

Viktor [21]

Some people naturally have more or less blemishes than others because of genes

6 0

3 years ago

Read 2 more answers

If the freezing point of an aqueous 0.10 m glucose solution is −x°c, what is the approximate freezing point of a 0.10 m nacl sol

maks197457 [2]

Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.

8 0

3 years ago