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Helen [10]
3 years ago
11

Which of the following aqueous solutions are good buffer systems?

Chemistry
1 answer:
Delicious77 [7]3 years ago
6 0

Answer:

0.25 M ammonium nitrate + 0.40 M ammonia  

0.39 M hypochlorous acid + 0.25 M potassium hypochlorite

0.12 M hydrofluoric acid + 0.14 M sodium fluoride  

Explanation:

A buffer consists of a weak acid and its salt or a weak base and its salt.

NH₃ is a weak base, and HClO and HF are weak acids.

B is wrong. HNO₃ is a strong acid.

C is wrong. KOH is a strong base.

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"sodium hydroxide is a strong base because it absorbs more" ____________ ions than weak bases.
Hunter-Best [27]
The answer is: absorbs more H (protons) ions.

- The Sodium hydroxide NaOH ionizes completely when dissolved in water.
- For every mole of sodium hydroxide that you dissolve you get 1 mole of hydroxide anions.
4 0
3 years ago
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Which element is most likely to gain an electron?
Oduvanchick [21]
B. Fluorine (F) is the right answer
3 0
3 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
A sample of Manganese (II) chloride has a mass of 19.8 grams before heating, and 12.6 grams after heating until all the water is
IRINA_888 [86]

Answer:

no. of water molecules associated to each molecule of MnCl_2 = 4

Explanation:

Mass of MnCl_2 before heating = 19.8 g

Mass of MnCl_2 after heating = 12.6 g

Difference in mass of MnCl_2 before and after heating

                                 = 19.8 - 12.6 = 7.2 g

Difference in mass corresponds to mass of water driven out.

Molar mass of water = 18 g/mol

No. of moles of water = \frac{7.2}{18} = 0.4\ mol

Mass of MnCl_2 obtained after heating is mass of anhydrous MnCl_2.

Mass of anhydrous MnCl_2 = 12.6 g

Molar mass of MnCl_2 = 125.9 g/mol

No. of mol of anhydrous MnCl_2 = \frac{125.9}{125.9} = 0.1\ mol

so,

0.1 mol of MnCl_2 have 0.4 mol of water

1 mol of MnCl_2 will have = \frac{0.4}{0.1} =4\ mol

Hence, no. of water molecules associated to each molecule of MnCl_2 = 4

5 0
3 years ago
For which of the following conversions does the value of the conversion factor depend upon the formula of the substance?
pochemuha

For which of the following conversions does the value of the conversion factor depend upon the formula of the substance?

The answer is C

3 0
3 years ago
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