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blondinia [14]
3 years ago
15

Potassium hydrogen phthalate, KHC8H5O4, or KHP, is used in many laboratories, including general chemistry laboratories, to stand

ardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A 0.3420-g sample of KHC8H5O4 reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH
Chemistry
1 answer:
Mamont248 [21]3 years ago
3 0

Answer:

Molarity of the sodium hydroxide solution is 0.04687 M.

Explanation:

Amount of potassium hydrogen phthalate = 0.3420 g

Moles of potassium hydrogen phthalate =\frac{0.3420 g}{204.2 g/mol}=0.001675 mol

KHC_8H_4O_4+NaOH\rightarrow NaKC_8H_4O_4+H_2O

According to reaction, 1 mole of potassium hydrogen phthalate reacts with 1 mole of sodium hydroxide, then 0.001675 moles pf potassium hydrogen phthalate will :

\frac{1}{1}\times 0.001675 mol=0.001675 mol of NaOH

Moles of sodium hydroxide = 0.001675 mol

Volume of sodium hydroxide solution =  35.73 mL = 0.03573 L  (1 mL = 0.001 L)

Molarity=\frac{Moles}{Volume (L)}  

Molarity of the sodium hydroxide solution :

=\frac{0.001675 mol}{0.03573 L}=0.04687 M

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alex41 [277]

Answer:

(<em>i) Concentrated HNO3 can be stored and transported in aluminium containers as it reacts with aluminium to form a thin protective oxide layer on the aluminium surface. This oxide layer renders aluminium passive. (ii) Sodium hydroxide and aluminium react to form sodium tetrahydroxoaluminate(III) and hydrogen gas.</em>

8 0
2 years ago
How many days does it take 16.Og of Gold-198 to decay to 1.0g? (each half-
forsale [732]

Answer:

10.8 days (3 sig.figs.)

Explanation:

All radioactive decay is 1st order decay defined by the expression A = A₀e^-kt

which is solved for time of decay (t) => t = ln(A/A₀) / -k

A = final weight = 1.0 gram

A₀ = initial weight = 16.0 grams

k = rate constant = 0.693/t(1/2) = 0.693/2.69 days = 0.258 days⁻¹

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4 0
2 years ago
What compound is formed when 2,2-dimethyloxirane (2-methyl-1,2-epoxypropane) is treated with ethanol containing sulfuric acid
vladimir1956 [14]

Answer:

2-ethoxy-2-methylpropan-1-ol

Explanation:

On this reaction, we have an "<u>epoxide"</u> (2-methyl-1,2-epoxypropane). Additionally, we have <u>acid medium</u> (due to the sulfuric acid H_2SO_4). The acid medium will produce the <u>hydronium ion</u> (H^+). This ion would be attacked by the oxygen of the epoxide. Then a <u>carbocation</u> would be produced, in this case, the most stable carbocation is the <u>tertiary one</u>. Then an <u>ethanol</u> molecule acts as a nucleophile and will attack the carbocation. Finally, a <u>deprotonation </u>step takes place to produce <u>2-ethoxy-2-methylpropan-1-ol</u>.

See figure 1

I hope it helps!

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3 years ago
Acetic acid and ethanol react to form ethyl acetate and water, like this:
ladessa [460]

Answer:

1.) Option C is correct.

The rate of reverse reaction is greater than zero, but equal to the rate of the forward reaction.

2) Option B is correct.

The rate of reverse reaction is Greater than zero, but less than the rate of the forward reaction.

3) Option C is correct.

The rate of reverse reaction is Greater than zero, and equal to the rate of the forward reaction.

4) Option A is correct.

How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium? Zero.

Explanation:

HCH,CO2(aq) + C2H5OH(aq) ⇌ C2H,CO2CH3(aq) + H2O

1) Before the main product is removed from the reaction setup, the chemical reaction is at equilibrium.

Chemical equilibrium is a state of dynamic equilibrium such that the concentration of the reactants and the products do not always remain the same but the rate of forward reaction always matches the rate of backward reaction.

2) When 246. mmol of C2HCO2CH3 are removed from the reaction mixture....

And when one of the factors involved in chemical equilibrium changes, Le Chatellier's principle explains that the system then adjusts to remedy this change and takes time to go back to equilibrium again.

When one of the species involved in the chemical reaction at equilibrium, is removed from the reaction mixture, the rate of reaction begins to favour that side of the reaction until equilibrium is re-established.

So, when 246 mmol of one of the products is removed, the response is to cause the rate of forward reaction to be favoured to produce more of products as there are fewer, and the rate of reverse reaction at this moment becomes less than the rate of forward reaction.

3) The rate of the reverse reaction when the system has again reached equilibrium

Like I said in (2) above, the reaction remedies this change in concentration of one of the products until equilibrium is re-established and when chemical equilibrium is re-established the rate of forward reaction once again matches the rate of backward reaction.

4) How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium?

By the time equilibrium is re-established, the system goes back to how it all was and the concentration of C2H5CO2CH3 goes back to the same as it was at the start of the reaction.

Hope this Helps!!!

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NemiM [27]

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