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Alexandra [31]
2 years ago
12

Acids and bases release what kind of ions in water?

Chemistry
2 answers:
Studentka2010 [4]2 years ago
6 0

Answer:

Explanation:

Acids and bases release what kind of ions in water?

Acids release hydrogen ions,  bases release hydroxide ions.

the cation H+ of acids consisting of a hydrogen atom whose electron has been transferred to the anion of the acid'

the monovalent anion OH− consisting of one atom of hydrogen and one of oxygen

— called also hydroxide ion

gizmo_the_mogwai [7]2 years ago
4 0

Answer:

"Acids produce hydrogen ions, H+, in water. Bases produce hydroxide ions, OH-."

Explanation:

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In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b
kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

Explanation:

Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)

The reaction is first order in [Co(NH_3)5Br]^{2+}:

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=0.100 M

a) Final concentration of [Co(NH_3)5Br]^{2+} after 19.0 hours= [A]

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}

[A]=0.065 M

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=x

Final concentration of [Co(NH_3)5Br]^{2+} after t = [A]=(100\%-69\%) x=31\%x=0.31x

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}

t = 185,902.06 s = \frac{185,902.06 }{3600} hour = 51.64 hours ≈ 52 hours

In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

6 0
3 years ago
An iron bar at 200c is placed in thermal contact with an identical iron bar at 120c in an isolated system
Step2247 [10]

Conduction: In the conduction, the heat is transferred from the hotter body to the colder body until the temperature on both bodies are equal.

In thermal equilibrium, there is no heat transfer as the heat is transferred till the temperature on the bodies are not same.

In the given problem, an iron bar at 200°C is placed in thermal contact with an identical iron bar at 120°C in an isolated system. After 30 minutes, the thermal equilibrium is attained. Then, the temperature on both iron bars are equal.Both iron bars are at 160°C in an isolated system.

But in an open system, the temperatures of the iron bars after 30 minutes would be less than 160°C. There will be heat lost to the surrounding. The room temperature is 25°C. There will be exchange of the heat occur between the iron bars and the surrounding. But It would take more than 30 minutes for both iron bars to reach 160°C because heat would be transferred less efficiently.

7 0
3 years ago
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fgiga [73]

Explanation:

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8 0
3 years ago
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What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the
lutik1710 [3]

Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

E°anode= -1.45 V

E°cell= -0.40-(-1.45) = 1.05 V

Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

Since log 1=0

Ecell= E°cell= 1.05 V

3 0
2 years ago
PLEASEEE HELPPPP ASAP!!
Goshia [24]

Atoms are made up of three subatomic particles called protons, neutrons, and electrons.

Protons and neutrons are located in the nucleus.

All protons have a positive charge.

All neutrons have no charge or are neutral.

Electrons orbit around the nucleus and have a negative charge.

8 0
3 years ago
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