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lawyer [7]
2 years ago
15

Which shows an isomer of the molecule below?

Chemistry
2 answers:
lakkis [162]2 years ago
5 0

Answer:

B should be ur answer!

Explanation:

Sorry if im wrong :( but have a good day or night

kobusy [5.1K]2 years ago
3 0

Answer:

B

Explanation:

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. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are
olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

K_{sp}\text{ of }Ag_2S=8\times 10^{-51}

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K

where,

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

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K = equilibrium constant or solubility product = 8\times 10^{-51}

Putting values in above equation, we get:

\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ

For the given chemical equation:

Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)

The equation used to calculate Gibbs free change is of a reaction is:  

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

The equation for the Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]

We are given:

\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ

Putting values in above equation, we get:

285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol

Hence, the standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

8 0
3 years ago
List the factor on which Ka of a weak acid depends​
lakkis [162]
I don’t know this, so I did research, so here! This is from a website called “Quora”, btw!
If you need more information, just go to this app by copying and pasting your question on the G o o g l e Search bar!

8 0
3 years ago
The vacoule is most like a ?
Feliz [49]
Animal’s and plant cells
3 0
3 years ago
Read 2 more answers
Question 4 point
Ronch [10]

Answer:

5393.4gH_2O

Explanation:

Hello there!

In this case, for the described chemical reaction, we can write:

CH_4+2O_2\rightarrow CO_2+2H_2O

Thus, by considering the 1:2 mole ratio of methane to water, and the molar mass of the latter (18.02 g/mol), the following is useful to calculate the mass of water that is produced:

149.65molCH_4*\frac{2molH_2O}{1molCH_4} *\frac{18.02gH_2O}{1molH_2O}

Which is equal to:

5393.4gH_2O

Which is not among the choices.

Regards!

5 0
3 years ago
Calculate the amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C.
qwelly [4]

Answer:

The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

Between heat and temperature there is a direct proportional relationship. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

  • c= 4.184 \frac{J}{g*C}
  • m= 32 g
  • ΔT= Tfinal - Tinitial= 22°C - 8°C= 14°C

Replacing:

Q= 32 g* 4.184 \frac{J}{g*C} *14 °C

Solving:

Q= 1,874.432 J

<u><em>The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J</em></u>

7 0
2 years ago
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