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Olegator [25]
3 years ago
11

Give the formula of each of the following (don't worry about subscripts or superscripts, example: HCO3- enter as HCO3-): the con

jugate acid of HSO3- H2SO3 the conjugate base of H2O the conjugate acid of NH3 the conjugate base of CH3NH2
Chemistry
2 answers:
azamat3 years ago
8 0

Answer:

See explanation below

Explanation:

In this case, for you to understand, let's do the acid base reaction, so you can see the conjugate base or acid.

For the case of the HSO3-, if you do the reaction:

HSO3- + H2O ------> H2SO3 + OH-

In this case, the negative charge, acts as a base, and took an hydrogen from the water. This forms the OH- and the conjugate acid of the HSO3-. The H2SO3 is called sulfurous acid.

For the case of H2SO3:

H2SO3 + H2O ---------> HSO3- + H3O+

The H2SO3 acts like acid, so the conjugate base would be the HSO3-

For the H2O:

2H2O --------> H3O+ + OH-

The water acts as base and acid at a time. In this case, when it acts like a base, took one hydrogen atom from the other molecule of water, and forms the H3O+. This would be the conjugate acid. The conjugate base would be the OH-

For the NH3:

NH3 + H2O --------> NH4+ + OH-

The atom of nitrogen in the NH3, took an atom of hydrogen to form the NH4+, which would be the conjugate acid.

For the CH3NH2:

CH3NH2 + H2O -------> CH3NH- + H3O+

The water took an atom of hydrogen, so, the water is the base and CH3NH2 is the acid, so the conjugate base is the CH3NH-

Regards

vovangra [49]3 years ago
3 0

Answer:

The relative conjugate acids and bases are listed below:

CH3NH2 → CH3NH3+

H2SO3→ HSO3-

NH3→ NH4+

Explanation:

In a Brønsted-Lowry acid-base reaction, a conjugate acid is the species resulting from a base accepting a proton; likewise, a conjugate base is the species formed after an acid has donated a hydrogen atom (proton).

To this end:

  • HSO3- is the conjugate acid of H2SO3 i.e sulfuric acid has lost a proton (H+)
  • NH4+ is the conjugate acid of NH3 i.e the base ammonia has gained a proton (H+)
  • OH- is the conjugate base of H20
  • CH3NH3+ is the conjugate base of the base CH3NH2 methylamine
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Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

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Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

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Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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3 years ago
Which compounds are classified as Arrhenius acids?
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Answer:The correct answer is option 4.

Explanation:

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BOH(aq)\rightarrow OH^-+B^+

HBr \& H_2SO_4 are Arrhenius acids because they form H^+ions in their respective aqueous solution.

HBr(aq)\rightarrow H^++Br^-

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Hence, the correct answer is option 4.

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Answer:

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By Charles's Law, under constant pressure the volume V of an ideal gas is proportional to its absolute temperature T (the one in degrees Kelvins.)

Alternatively, consider the ideal gas law:

\displaystyle V = \frac{n \cdot R}{P}\cdot T.

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Therefore the volume of the gas is proportional to its absolute temperature.

Either way,

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Convert the two temperatures to degrees Kelvins:

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Apply Charles's Law:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 6.24\;L \times \frac{328.15\; K}{298.15\;K} = 6.87\;L.

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