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Bess [88]
2 years ago
10

Since vinegar is 5% acetic acid and 95% water, find the mole ratio of acetic acid to water in 100 g of vinegar? Hint: Instead of

determining % composition of elements you have % composition of molecules.

Chemistry
1 answer:
Oksanka [162]2 years ago
7 0

The mole ratio of acetic acid to water in 100 g of vinegar is 0.015 : 0.985.

<h3>What is the mole ratio of acetic acid to water in 100 g of vinegar?</h3>

The mole ratio of acetic acid to water in 100 g of vinegar is determined from their percentage composition.

The percentage composition of acetic acid and water in vinegar is 5% acetic acid and 95% water.

In 100 g of vinegar, there are 5 g of acetic acid and 5 g of water.

Moles = mass/molar mass

molar mass of acetic acid = 62 g/mol

molar mass of water = 18 g/mol

moles of vinegar = 5/62 = 0.08

moles of water = 95/18 = 5.28

total moles = 5.36

Mole ratio of vinegar to water = 0.08/5.36 : 5.28/5.36

Mole ratio of vinegar to water = 0.015 : 0.985

In conclusion, the mole ratio is determined from the percentage composition of acetic acid and water in vinegar.

Learn more about mole ratio at: brainly.com/question/19099163

#SPJ1

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• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

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Derivative for tan(x)
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Answer:

  • sec² (x)

Explanation:

Use the trigonometric ratio definition of the tangent function and the quotient rule.

  • tan(x) = sin(x) / cos(x)

Quotient rule: the derivative of a quotient is:

  • [the denominator × the derivative of the numerator less the numerator × the derivative of the denominator] / [denominator]²

  • (f/g)' = [ g×f' - f×g'] / g²

So,

  • tan(x)' = [ sin(x) / cos(x)]'
  • [ sin(x) / cos(x)]' = [ cos(x) sin(x)' - sin(x) cos(x)' ] / [cos(x)]²

                                   = [ cos(x)cos(x) + sin(x) sin(x) ] / [ cos(x)]²

                                   = [ cos²(x) + sin²(x) ] / cos²(x)

                                   = 1 / cos² (x)

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Answer:

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Width of the water bath, W= 0.810  m

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Height of the water in water bath, h= 0.740 m - 2.57 inches

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