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timurjin [86]
3 years ago
8

artaric acid, H2C4H4O6, has two acidic hydrogens. The acid is often present in wines and precipitates from solution as the wine

ages. A solution containing an unknown concentration of the acid is titrated with NaOH. It requires 21.65 mL of 0.3500 M NaOH solution to titrate both acidic protons in 50.00 mL of the tartaric acid solution. You may want to reference (Pages 149 - 153) Section 4.6 while completing this problem. Part A Write a balanced net ionic equation for the neutralization reaction. Express your answer as a chemical equation including phases.
Chemistry
1 answer:
elena55 [62]3 years ago
6 0

Answer:

2H⁺(aq) + + 2OH⁻(aq) → 2H₂O(l)

Explanation:

A acid that has two acidic hydrogens requires to moles of NaOH per mole of acid producing the salt of the acid and water.

If the acid is tartaric acid, H₂C₄H₄O₆, the reaction with NaOH is:

<em>H₂C₄H₄O₆(aq) + 2NaOH(aq) → 2H₂O(l) + 2NaC₄H₄O₆(aq)</em>

Now, ions in solution in the reaction are:

2H⁺(aq) + C₄H₄O₆²⁻(aq) + 2Na⁺(aq) + 2OH⁻(aq) → 2H₂O(l) + 2Na⁺(aq) +C₄H₄O₆²⁻(aq)

Thus, net ionic equation (Lisiting only the ions involved in the reaction) is:

<h3>2H⁺(aq) + + 2OH⁻(aq) → 2H₂O(l) </h3>

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3 years ago
Suppose 50.0g of silver nitrate is reacted with 50g of hydrochloric acid producing silver chloride and a mixture of other produc
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Answer:

53.6 grams of silver chloride was produced.

Explanation:

AgNO_3+HCl+\rightarrow AgCl+HNO_3

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This also means that total mass on the reactant side must be equal to the total mass on the product side.

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Mass of hydrogen chloride = 50.0 g

Mass of silver chloride = x

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Mass of silver nitrate + Mass of hydrogen chloride =

                             Mass of silver chloride + Mass of  nitric acid

[te]50.0 g+50.0 g=x+46.4 g[/tex]

x=50.0 g+50.0 g - 46.4 g = 53.6 g

53.6 grams of silver chloride was produced.

8 0
3 years ago
When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
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Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
3 years ago
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