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11 months ago

determine whether the aqueous solutions of each of the following salts will be acidic, basic, or neutral

Chemistry

1 answer:

eduard11 months ago

6 0

KCl is neutral because KOH is a strong base and HCl is a strong acid

To determine if the salt is neutral, acidic or basic, determine how the salt was made, or from which acid and base it was derived. Once we have determined how the salt was made, determine if the acid and base are both strong, weak or one of each. If both are strong (e.g. NaOH + HCl), then the salt (NaCl) is NEUTRAL. If the base is strong and the acid is weak (KOH + HCN), then the salt (KCN) is BASIC. IF the base is weak and the acid is strong (NH3 + HCl), then the salt (NH4Cl) is ACIDIC. If both the base and the acid are weak (NH3 + CH3COOH), then then salt (CH3COONH4) will have a pH determined by which is stronger (the acid or the base) and this will depend on the Ka of the weak acid compared to the Kb of the weak base.

KCl is neutral because KOH is a strong base and HCl is a strong acid NaCN is generally basic because NaOH is a strong base and HCN is a very week acid. NH4NO3 is usually acid because HNO3 is a strong acid and NH4OH is a week base

CH3COONH4 is exactly neutral because both NH4OH ad CH3COOH are respectively week base and week acid. Moreover, the Ka of acetic acid is exactly equal to the Kb of NH4OH

To know about strong base.

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Explanation:

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2 years ago

Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a b

Alinara [238K]

Answer:

Explanation:

The cell reaction properly written is shown below:

Cu|Cu²⁺ $_{aq}$ || Ag⁺ $_{aq}$ | Ag

From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.

Oxidation half:

Cu $_{s}$ ⇄ Cu²⁺ $_{aq}$ + 2e⁻

At the anode, oxidation occurs.

Reduction half:

Ag⁺ $_{aq}$ + 2e⁻ ⇄ Ag $_{s}$

At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

Cu $_{s}$ ⇄ Cu²⁺ $_{aq}$ + 2e⁻

Ag⁺ $_{aq}$ + e⁻ ⇄ Ag $_{s}$

we multiply the second reaction by 2 to balance up:

2Ag⁺ $_{aq}$ + 2e⁻ ⇄ 2Ag $_{s}$

The net reaction equation:

Cu $_{s}$ + 2Ag⁺ $_{aq}$ + 2e⁻⇄ Cu²⁺ $_{aq}$ + 2e⁻ + 2Ag $_{s}$

We then cancel out the electrons from both sides since they appear on both the reactant and product side:

Cu $_{s}$ + 2Ag⁺ $_{aq}$ ⇄ Cu²⁺ $_{aq}$ + 2Ag $_{s}$

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2 years ago

An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at

Yakvenalex [24]

Answer:

68.3g

Explanation:

1. Word equation:

mercury(II) oxide → mercury + oxygen

2. Balanced molecular equation:

2HgO → 2Hg + O₂(g)

3. Mole ratio

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

4. Calculate the number of moles of O₂(g)

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

5. Calculate the number of moles of HgO

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

6. Convert to mass

mass = # moles × molar mass

molar mass of HgO: 216.591g/mol

mass = 0.315mol × 216.591g/mol = 68.3g

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2 years ago

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Mnenie [13.5K]

Answer:

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2 years ago

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Marat540 [252]

When you immerse an ionic compound in water, the ions are attracted to the water molecules, each of which carries a polar charge. If the attraction between the ions and the water molecules is great enough to break the bonds holding the ions together, the compound dissolves.

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2 years ago