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ohaa [14]
3 years ago
5

A common cooking vinegar is 5.0% acetic acid (ch3cooh. how many molecules of acetic acid are present in 41.9 g vinegar?

Chemistry
1 answer:
UkoKoshka [18]3 years ago
6 0
To calculate the number of molecules ch3cooh in a given amount of vinegar, we first need to determine the amount of ch3cooh in the solution by multiplying the percent <span>ch3cooh with the sample amount. Then, convert g to moles and use avogadro's number. We do as follows:

41.9 g vinegar solution ( .05 g </span><span>ch3cooh / g vinegar solution ) (1 mol / 60.05 g ) ( 6.022x10^23 g / 1 mol ) = 2.1x10^22 molecules </span><span>ch3cooh</span>
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When a candle is lit and allowed to burn for 15 minutes, some wax drips and collects at the base of the candle, and the candle b
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The missing section of the candle burns away and evaporates while some drips and collects at the base.

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4 years ago
If water was used to rinse the burette used to dispense the hydrochloric acid solution, how would this affect
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The concentration of hydrochloric acid would be estimated to be less.

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3 years ago
How can ionic bonds and naming be used in a real life example ?
Stells [14]

Answer:

Melting snow more efficiently in winters, understanding the components of mineral water

Explanation:

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8 0
3 years ago
A 25.0 mL sample of H3PO4 requires 50.0 mL of 1.50 M NaOH for complete neutralization. What is the molarity of the acid?
Kamila [148]

Answer:

The molarity of acid is 3 M.

Explanation:

Given data:

Volume of H₃PO₄ = 25 mL

Volume of NaOH = 50 mL

Molarity of NaOH = 1.50 M

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Solution:

Formula:

M₁V₁  =  M₂V₂

M₁ = M₂V₂ / V₁

M₁ = 1.50 M ×50 mL / 25 mL

M₁ = 75 M. mL / 25 mL

M₁ = 3 M

The molarity of acid is 3 M.

8 0
4 years ago
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