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The pH of a vinegar solution is 4.15. What is the H+ concentration of the solution

marishachu [46]

The pH of a vinegar solution is 4.15. To find the H+ concentration of the solution use the following equation -log(H+)=pH. Insert the pH into the equation to get, -log(H+) = 4.15 Rearrange the equation to get, 10^(-4.15) = H+ Finally, you can solve for H+. The hydrogen ion concentration of the vinegar solution is .0000708 M.

3 0

3 years ago

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An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam

MAXImum [283]

Explanation:

The given data is as follows.

Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence, moles of NaOH in 97 ml = $\frac{0.1090 \times 97}{1000}$

= $10.573 \times 10^{-3} mol$

Moles of HCl in 21.00 ml = $\frac{0.2060 \times 21}{1000}$

= $4.326 \times 10^{-3}$ mol

Therefore, total moles of NaOH that reacted are as follows.

$10.573 \times 10^{-3} mol$ - $4.326 \times 10^{-3} mol$

= $6.247 \times 10^{-3}$ mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

$\frac{6.247 \times 10^{-3}}{3}$

= $2.082 \times 10^{-3}$ mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

$2.082 \times 10^{-3} mol \times 40 g/mol$

= $83.293 \times 10^{-3}$ g

= 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore, 40 g NaOH = 122 g benzoic acid

So, 0.0832 g NaOH = $\frac{122 g}{40 g} \times 0.0832 g$

= 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

$\frac{0.253 g}{0.3471 g} \times 100$

= 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0

3 years ago

For the reaction N2(g) + 2H2(g) → N2H4(l), if the percent yield for this reaction is 100.0%, what is the actual mass of hydrazin

barxatty [35]

Answer:

53.6 g of N₂H₄

Explanation:

The begining is in the reaction:

N₂(g) + 2H₂(g) → N₂H₄(l)

We determine the moles of each reactant:

59.20 g / 28.01 g/mol = 2.11 moles of nitrogen

6.750 g / 2.016 g/mol = 3.35 moles of H₂

1 mol of N₂ react to 2 moles of H₂

Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.

2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine

Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄

Let's convert the moles to mass:

1.67 mol . 32.05 g/mol = 53.6 g

4 0

3 years ago

Cytochromes are critical participants in the electron transport chains used in photosynthesis and cellular respiration. How do c

goblinko [34]

Answer:

4) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.

Explanation:

The cytochromes are proteins that contain heme prosthetic groups. Cytochromes undergo oxidation and reduction through loss or gain of a single electron by the iron atom in the heme of the cytochrome:

$Cytochrome-Fe²⁺ ⇄ cytochrome-Fe³⁺-e⁻$

The reduced form of ubiquinone (QH₂), an extraordinarily mobile transporter, transfers electrons to cytochrome reductase, a complex that contains cytochromes b and c₁, and a Fe-S center. This second complex reduces cytochrome c, a water-soluble membrane peripheral protein. Cytochrome c, like ubiquinone (Q), is a mobile electron transporter, which is transferred to cytochrome oxidase. This third complex contains the cytochromes a, a₃ and two copper ions. Heme iron and a copper ion of this oxidase transfer electrons to O₂, as the last acceptor, to form water.

Each transporter "downstream" is more electronegative than its neighbor "upstream"; oxygen is located in the inferior part of the chain. Thus, the electrons fall in an energetic gradient in the electron chain transport to a more stable localization in the electronegative oxygen atom.

7 0

3 years ago

Draw the structure(s) of all of the possible monochloro derivatives of 2,4-dimethylpentane, c7h15cl.

pishuonlain [190]

The monochloroderivatives will be obtained by substituting chemically non equivalent hydrogen with chlorine atom, one by one

So the possible monochloro derivatives of 2,4-dimethylpentane (figure 1) are shown in figure (2)

7 0

3 years ago