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miss Akunina [59]
3 years ago
10

Which reaction displays an example of an Arrhenius base? NaOH(s) mc005-1.jpg Na+(aq) + OH–(aq) HCl(g) + H2O(l) mc005-2.jpg H3O+(

aq) + Cl–(aq) CH3COOH(aq) + H2O(l) mc005-3.jpg H3O+(aq) + CH3COO–(aq) NH3(aq) + HC2H3O2(aq) mc005-4.jpg NH4+(aq) + C2H3O2–(aq)
Chemistry
2 answers:
dangina [55]3 years ago
8 0

The reaction which displays an   example of Arrhenius  base is

 NaOH(s) →  Na +(aq)  + OH- (aq)

 <u><em>Explanation</em></u>

  •  according to Arrhenius a  base is a substance that dissociate    in water to  hydroxide ions ( OH-).
  •  The  reaction of NaOH(s) → Na+(aq)   +OH-(aq   is an example of Arrhenius base since  NaOH  dissociate in water  to form hydroxide ions (OH-)
vivado [14]3 years ago
7 0

Answer:

NaOH(s) →  Na +(aq)  + OH- (aq)

Explanation:

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What amounts of sodium benzoate would be required to prepare 2.5L of 0.35M benzoic buffer solution with a pH of 6.10? Ka of benz
slava [35]

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

3 0
3 years ago
9. Write the number in place value table of 820245<br>​
Burka [1]

Answer:

8----hundred thousands

2----ten thousands

0----thousands

2----hundreds

4----tens

5---ones

place value

 8 = 800,000

 2 = 20,000

 0 = 0,000

2 = 200

4 =40

5 =5

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3 years ago
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What is a process of reacting an acid with a base until an equivalence point is reached
Brrunno [24]
TITRATION is the process of reaching equilibrium between acids and bases.
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3 years ago
Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3
vichka [17]

Answer:

0.13 m of Mn(NO_3)_2 → Highest boiling point

0.19 m of AgNO_3 → Second  Highest boiling point

0.17 m of CrSO_4 → Third highest boiling point

0.31 m Sucrose (nonelectrolyte)  → Lowest boiling point

Explanation:

Elevation in boiling is given by :

\Delta T_b=i\times k_b\times m

Where :

i = van't Hoff factor

k_b= Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of AgNO_3

AgNO_3\rightarrow Ag^++NO_3^{-}

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

\Delta T_b=2\times k_b\times 0.19 m

\Delta T_b=0.38 m\times k_b

2) 0.17 m of CrSO_4

CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

\Delta T_b=2\times k_b\times 0.17 m

\Delta T_b=0.34 m\times k_b

3) 0.13 m of Mn(NO_3)_2

Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

\Delta T_b=3\times k_b\times 0.13 m

\Delta T_b=0.39 m\times k_b

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

\Delta T_b=1\times k_b\times 0.31 m

\Delta T_b=0.31 m\times k_b

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b

0.13 m of Mn(NO_3)_2 → Highest boiling point

0.19 m of AgNO_3 → Second  Highest boiling point

0.17 m of CrSO_4 → Third highest boiling point

0.31 m Sucrose (nonelectrolyte)  → Lowest boiling point

6 0
3 years ago
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