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IgorLugansk [536]
2 years ago
12

[ the (H +) of a solution with pH 4.0 is​

Chemistry
1 answer:
konstantin123 [22]2 years ago
7 0

Answer:

In this way, pH is determined by hydrogen-ion concentration. In the case of a neutral solution, [H+]=10-7 , which we call a pH of 7. This means, for example, that a hydrogen-ion concentration of a solution with a pH of 4 is 10-4mol/l, meaning it contains 0.0001 mol of hydrogen ions in a solution of 1 liter. I explained everything I hope it is enough

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Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu
nydimaria [60]

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^0_{[Pb^{2+}/Pb]}=-0.13V

E^0_{[Cu^{2+}/Cu]}=+0.34V

E^0_{cell}=E^0_{cathode}-E^0_{anode}

E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}

E^0_{cell}=-0.13V-(0.34V)=-0.47V

Now we have to calculate the standard Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole

Therefore, the standard Gibbs free energy is +91 kJ/mole

6 0
2 years ago
7. The youngest part of the ocean floor is found ____.
aev [14]

Answer:

The answer is Near The Ocean Ridges.

5 0
2 years ago
Based on the solubility observations, which of the following pairs of cations could be distinguished by the addition of sodium c
Licemer1 [7]

Based on the solubility observations, barium & aluminum could be distinguished by the addition of sodium chloride to the solutions.

<h3>What happens when NaCl is added to a solution?</h3>
  • The ionic link that held sodium and chloride ions together is broken when water molecules force the ions apart.
  • The sodium and chloride atoms are encircled by water molecules after the salt compounds are separated. After that, the salt dissolves and forms a homogenous solution.
  • In order to keep patients from dehydrating, sodium chloride, an important nutrient, is employed in healthcare. It is employed as a spice to improve flavor and as a food preservative. Additionally, sodium chloride is employed in the production of polymers and other goods. Additionally, it is used to de-ice sidewalks and roadways.
  • Adding water to sodium chloride results in a physical change because no new product is created.

Learn more about sodium chloride added to a solution refer to :

brainly.com/question/28092739

#SPJ4

7 0
10 months ago
HELP IM TIMEDDDDDDDDDDDD
nadya68 [22]

Answer:

ethier a dessert or a plains

though plains can get rain in the summer

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3 0
2 years ago
Read 2 more answers
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
2 years ago
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