What is the process of solvation of ionic solids in water?

Answers

Answer 1
Answer: The solvation is the process that causes the dissolution of ionic solids, because the energy released compensates for the energy required to break the Crystal network. Occurs only with polar solvents. The solvation in which the solvent is water is called hydration.

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What is a covalent bond?​

Answers

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding. (hope this helps.)

Answer:

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

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H2(g) + CO2(g) + 10 kcal H2O(g) + CO(g)In another experiment involving the above reaction at 1,000°C, the equilibrium concentrations were found to be:

H2(g) = 2.0 moles per liter
H2O(g) = 4.0 moles per liter
CO2(g) = 5.0 moles per liter
CO(g) = 4.0 moles per liter

What is the equilibrium constant, Keq, for this reaction?

Answers

The equilibrium contant for the reaction is 1.6. Solution: [H2]=2.0 mol/L [CO2]= 5.0 mol/L [H2O]= 4.0 mol/L [CO]= 4.0 mol/L K = [H2O][CO] / [H2][CO2] K = (4.0)(4.0) / (2.0)(5.0) K = 1.6 Ans [ =1.6 ]

In terms of energy and entropy, systems in nature tend to undergo changes toward(1) higher energy and higher entropy
(2) higher energy and lower entropy
(3) lower energy and higher entropy
(4) lower energy and lower entropy

Answers

In terms of energy and entropy, systems in nature tend to undergo changes toward lower energy and higher entropy. Entropy is important as it describes the flow and distribution of energy. For a process to to occur spontaneously, it is a necessary condition that the entropy of the system undergoing the process to increase. If the entropy decreases, then that process cannot occur spontaneously. It needs some work/ energy to be pumped in in order for the process to occur
In terms of energy and entropy, systems in nature tend to undergo changes toward lower energy and higher entropy.

If i add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be ?

Answers

Taking into account the definition of dilution, the molarity of the diluted solution is 0.125 M.

Dilution

When it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

  • Ci: initial concentration
  • Vi: initial volume
  • Cf: final concentration
  • Vf: final volume

Final molarity

In this case, you know:

  • Ci= 0.15 M
  • Vi= 125 mL
  • Cf= ?
  • Vf= 125 mL + 25 mL= 150 mL

Replacing in the definition of dilution:

0.15 M× 125 mL= Cf× 150 mL

Solving:

(0.15 M× 125 mL)÷ 150 mL= Cf

0.125 M= Cf

In summary, the molarity of the diluted solution is 0.125 M.

Learn more about dilution:

brainly.com/question/6692004

brainly.com/question/16343005

brainly.com/question/24709069

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Answer:

0.125 M

Explanation:

M₁V₁ = M₂V₂

125 mL / 1000 = 0.125 L = V₁

25 + 125 = 150 mL / 1000 = 0.150 L = V₂

0.15 M = M₁

M₁ × V₁ / V₂ = M₂

(0.15) × (0.125) / (0.150) = 0.125 M

0.125 M is the answer

How is helium which is produced by the conversion of hydrogen into energy related to the solar core

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The sun's core is where Hydrogen atoms produce energy as they convert to molecules of Helium. Helium that is produced by the conversion of hydrogen into energy is related to the solar core in the sense that the conversion process produces energy captured in the core. I hope this helps you. 

Answer : The phenomenon called as Nuclear Fusion is observed in the sun's core or solar core, where 4 atoms of Hydrogen molecule combines to give two atoms of Helium. It is due to fusing of two Hydrogen atoms to form one Helium molecule. In the core of sun this fusion process occurs, changing of mass of Hydrogen atom into Helium releases lot of energy in the form of heat and light.


The reaction that occurs inside the solar core is given below;


 4 H^(1)_(1) ----> He^(4)_(2)

Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions. Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) S (s) (nonspontaneous)? Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq) (nonspontaneous)? 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq) (spontaneous?) Are these correct?

Answers

A electrochemical reaction is said to be spontaneous, if E^(0) cell is positive. 

Answer 1:
Consider reactionNi^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
    
S^(2-)/S // Ni^(2+)/Ni

Hence, E^(0)cell = E^(0) Ni^(2+/Ni) - E^(0) S/S^(2-)
we know that, {E^(0) Ni^(2+)/Ni = -0.25 v
and {E^(0) S/ S^(2-) = -0.47 v

Therefore, E^(0) cell = - 0.25 - (-0.47) = 0.22 v

Since,  E^(0) cell is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2: 
Consider reactionPb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

The cell representation of above reaction is given by;
    H_(2) / H^(+) // Pb^(2+) /Pb

Hence, E^(0)cell = E^(0) Pb/Pb^(2+) - E^(0) H_(2)/H^(+)
we know that, {E^(0) Pb^(2+)/Pb = -0.126 v
and {E^(0) H_(2)/ H^(+) = -0 v

Therefore, E^(0) cell = - 0.126 - 0 = -0.126 v

Since,  E^(0) cell is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3
Consider reaction2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)

The cell representation of above reaction is given by;
    Cr/Cr^(2+) // Ag^(+)/Ag

Hence, E^(0)cell = E^(0) Ag^(+)/Ag - E^(0) Cr/Cr^(2+)
we know that, {E^(0) Ag^(+)/Ag = -0.22 v
and {E^(0) Cr/ Cr^(2+) = -0.913 v

Therefore, E^(0) cell = - 0.22 - (-0.913) = 0.693 v

Since,  E^(0) cell is positive, hence cell reaction is spontaneous

Answer: Ni^(2+)(aq)+S^(2-)(aq)\rightarrow Ni(s)+S(s)  : non spontaneous

Pb^(2+)(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)  : non spontaneous

2Ag^(+)(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^(2+)(aq)  : spontaneous

Explanation:

a) Ni^(2+)(aq)+S^(2-)(aq)\rightarrow Ni(s)+S(s)

Here S undergoes oxidation by loss of electrons, thus act as anode. Ni undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_(cathode)- E^0_(anode)

Where both E^0 are standard reduction potentials.

E^0_([Ni^(2+)/Ni])=-0.25V

E^0_([S^(2-)/S])=0.407VV

E^0=E^0_([Ni^(2+)/Ni])- E^0_([S^(2-)/S])

E^0=-0.25-(0.407V)-0.657V

As value of E^0 is negative, the reaction is non spontaneous.

b)Pb^(2+)(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)

Here Hydrogen undergoes oxidation by loss of electrons, thus act as anode. Pb undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_(cathode)- E^0_(anode)

Where both E^0 are standard reduction potentials.

E^0_([Pb^(2+)/Pb])=-0.13

E^0_([H^(+)/H_2])=0V

E^0=E^0_([Pb^(2+)/Pb])- E^0_([H^(+)/H_2])

E^0=-0.13-(0V)=-0.13V

As value of E^0 is negative, the reaction is non spontaneous.

c) 2Ag^(+)(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^(2+)(aq)

Here Cr undergoes oxidation by loss of electrons, thus act as anode. Ag undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_(cathode)- E^0_(anode)

Where both E^0 are standard reduction potentials.

E^0_([Ag^(+)/Ag])=+0.80V

E^0_([Cr^(2+)/Cr])=-0.913V

E^0=E^0_([Ag^(+)/Ag])- E^0_([Cr^(2+)/Cr])

E^0=+0.80-(-0.913V)=1.713V

As value of E^0 is positive, the reaction is spontaneous.