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svet-max [94.6K]

3 years ago

13

Rank each of the following gases in order of increasing urms assuming equivalent amounts and all gases are at the same temperatu

re and pressure where 1 has the lowest urms and 4 has the highest urms.
a. Gas 1 : H2S
b. Gas: He
c. Gas 3: NF3
d. Gas 4: H2O

1 answer:

anzhelika [568]3 years ago

5 0

the answer is option c

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Answer:

Benzoic acid: 1.288g

Sodium benzoate: 124.48g

Explanation:

Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:

pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)

<em>Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.</em>

As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:

2.5L ₓ (0.35mol / L) = 0.875moles of buffer.

And you can write:

0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)

Replacing (2) in (1)

pH = pKa + log [C7H5O2⁻] / [HC7H5O2]

6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]

1.913 = log [C7H5O2⁻] / [HC7H5O2]

81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]

81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]

82.846 [HC7H5O2] = 0.875mol

<h3>[HC7H5O2] = 0.01056 moles</h3>

And moles of the benzoate, [C7H5O2⁻]:

[C7H5O2⁻] = 0.875mol - 0.01056mol =

<h3>[C7H5O2⁻] = 0.8644mol</h3><h3 />

Using molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:

Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g

Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g

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Explanation:

Grignard reagent is a compound formed by the reaction of halide of alkyl or alkene with magnesium metal. This makes the compound more nucleophilic. Thus the availability of an electrophile around it will evoke a quick reaction. The presence of water in an acidic condition will generate hydroxonium ions which are highly electrophilic. There will therefore be an introduction of hydrogen ions to the grignard reagent and which will displace the MgBr leading to the formation of alkane.

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