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3 years ago

Which of the following is a base? AgO Ca(OH)2 HF NaCl

Chemistry

2 answers:

nikdorinn [45]3 years ago

6 0

<u>Answer:</u> The correct answer is $Ca(OH)_2$

<u>Explanation:</u>

A base is defined as the substance which looses hydroxide ions $(OH^-)$ when dissolved in water.

An acid is defined as the substance which looses hydrogen ions $(H^+)$ when dissolved in water.

From the given options:

AgO and NaCl are the salts. Both are ionic salts which are formed from the reaction of a metal and an non-metal.

HF is an acid because it released $H^+$ ions when dissolved in water. The equation for the ionization of HF follows:

$HF\rightleftharpoons H^++F^-$

$Ca(OH)_2$ is a base because it released $OH^-$ ions when dissolved in water. The equation for the ionization of $Ca(OH)_2$ follows:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+OH^-$

Hence, the correct answer is $Ca(OH)_2$

givi [52]3 years ago

3 0

Answer:

Ca(OH)

Explanation:

mark brainliest please :)

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The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other plati

Alinara [238K]

Here we have to identify the metal powder by the given disposal.

The identification of Zinc can be done by 1 m nitric acid (HNO₃) and Ni(NO₃)₂ which will produce hydrogen gas by reaction and displacement reaction as shown below.

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

The identification of lead can be done by the reaction with 1 m nitric acid (HNO₃) which produces lead nitrate.

The reaction is-

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

The identification of platinum can be done by the reaction with all the given disposal as it will not react with any of the compound.

1. Identification of Zinc (Zn):

(a) Zn metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than zinc.

Zn + NaNO₃ = No reaction

(b) Zn metal will react with 1 m HNO₃ to form hydrogen gas. The reaction is:

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

(c) Zn will react with nickel nitrate [Ni(NO₃)₂] because it may only cause displacement reaction the reduction potential of Zn²⁺/Zn (-0.76) is less than that of Ni²⁺/Ni (-0.23). Thus the reaction will be:

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

2. Identification of lead (Pb):

(a) Pb metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pb.

Pb + NaNO₃ = No reaction

(b) Pb reacts with HNO₃ to form lead nitrate. The reaction is:

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

(c) The standard reduction potential of Pb²⁺/Pb is more than nickel Ni²⁺/Ni thus there will be no reaction between Pb and NI(NO₃)₂.

Pb + Ni(NO₃)₂ = No reaction.

3. Identification of platinum (Pt)

(a) Pt metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pt.

Pt + NaNO₃ = No reaction.

(b) The standard reduction potential of Pt²⁺/Pt is so high (+1.188) thus there will be no reaction with HNO₃.

Pt + HNO₃ = No reaction

(c) The standard reduction potential of Pt²⁺/Pt is more than nickel Ni²⁺/Ni thus there will be no reaction between Pt and Ni(NO₃)₂.

Pt + Ni(NO₃)₂ = No reaction.

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3 years ago

A chemical bond is a strong attraction between two or more atoms. Can you identify the properties of different types of chemical

kakasveta [241]

Answer:

1. Ionic bond.

2.Ions

3. Covalent

4.Single bond

5. Double bond

6. Polar bond

7.Non polar bond

8. Hydrogen bond

5 0

3 years ago

Read 2 more answers

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

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Which of the following reactions involves a single compound producing two or more simpler substances?

castortr0y [4]

Answer:

Decomposition or cracking

Explanation:

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B and C are the products

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nevsk [136]

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3 years ago