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77julia77 [94]
2 years ago
5

What is the difference between applied and pure chemistry?

Chemistry
1 answer:
My name is Ann [436]2 years ago
5 0
I hope this helps.
have a great day.

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The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−
barxatty [35]

Answer:

The reaction will be  non spontaneous at these concentrations.

Explanation:

AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)

Expression for an equilibrium constant K_c:

K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]

Solubility product of the reaction:

K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}

Reaction between Gibb's free energy and equilibrium constant if given as:

\Delta G^o=-2.303\times R\times T\times \log K_c

\Delta G^o=-2.303\times R\times T\times \log K_{sp}

\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]

\Delta G^o=69,117.84 J/mol=69.117 kJ/mol

Gibb's free energy when concentration [Ag^+] = 1.0\times 10^{-2} M and [Br^-] = 1.0\times 10^{-3} M

Reaction quotient of an equilibrium = Q

Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}

\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)

\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])

\Delta G=40.588 kJ/mol

  • For reaction to spontaneous reaction:  \Delta G.
  • For reaction to non spontaneous reaction:  \Delta G>0.

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0
3 years ago
The combination of sulfur dioxide, nitrogen oxide, and atmospheric moisture creates
sweet-ann [11.9K]
This combination creates acid rain 
8 0
3 years ago
Read 2 more answers
H2CO3(aq) + H200 H30 (aq) + HCO3 (aq).
timofeeve [1]

Answer:

K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}

Explanation:

Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:

  • an equilibrium constant is, first of all, a fraction;
  • in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
  • in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
  • each concentration should be raised to the power of the coefficient in the balanced chemical equation;
  • only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.

Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:

K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}

4 0
3 years ago
In sediments and waterlogged soil, dissolved O2 concentrations are so low that the microorganisms living there must rely on othe
kakasveta [241]

Answer:

1) SO₄ ²⁻ : (+6)

  H₂S : (-2)

Explanation:

a) <u>Sulfate reducers</u> are widespread in muds and other sediments, water-logged soils, etc., environments that contain SO₄ ²⁻ and become anoxic as a result of microbial decomposition.

Sulfate (SO₄ ²⁻), the most oxidized form of sulfur (+6), <u>is reduced</u> by these

sulfate-reducing bacteria. The end product of sulfate reduction is hydrogen sulfide, H₂S, (oxidation number -2) an important natural product that participates in many biogeochemical processes. The H₂S they generate is responsible for the pungent smell (like that of rotten eggs) often encountered near coastal ecosystems. When sulfate-reducing bacteria grow, the H₂S formed from SO₄ ²⁻ reduction combines with the ferrous iron to form black, insoluble ferrous sulfide, which is not toxic. This is important for the conservation of the environment.

b) The net ionic equation under acidic conditions is:

              4 H₂ + SO₄²⁻ + H⁺ → HS⁻ + 4 H₂O

    Global reaction:  SO₄²⁻ + 2H⁺ → H₂S + O₂

3 0
2 years ago
9474 millimeters to centimeters using scientific notation showing work
puteri [66]

Answer:

9.474 x 10^2

Explanation:

ok. first you have to get the value in the required unit so 9474mm/(10mm/cm) = 947.4 so scientific notation states that the number must be raised to any power of an integer and the value of the number being raised must be less than than 10 and more than or equal to 1

so it must have one digit in front so.. 947.4 becomes 9.474 and because you move 2 places to the left, ur power is positive 2

and proof 10^2 is 100 so multiply 9.474 by 100 and u will get 947.4 cm which is also 9474 mm

8 0
3 years ago
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