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Misha Larkins [42]
3 years ago
13

A sample of mass 6.814 grams is added to another sample weighing 0.08753 grams.

Chemistry
1 answer:
Feliz [49]3 years ago
3 0

Answer:

17.5609g

Explanation:

According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;

6.814 + 0.08753 = 6.90153grams

Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3

= 6.90153/3

= 2.30051grams.

One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams

Therefore, the final mass is 17.5609grams

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Answer:

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The model of the atom supported by Bohr's hydrogen experiment is the Rutherford's model of the atom.

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For the reaction C2H4(g) + H2O(g) --&gt; CH3CH2OH(g)
Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

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

where,

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

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

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