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valentina_108 [34]
3 years ago
10

Water (H2O) is a liquid at room temperature. Ammonia (NH3) is a gas at room temperature. Both are polar covalent compounds. Whic

h compound most likely has the strongest intermolecular forces?
a. water
b ammonia
Chemistry
2 answers:
Nat2105 [25]3 years ago
8 0

the answer is: A. Water

Tanzania [10]3 years ago
5 0

Answer:

Water has the strongest inter-molecular forces.

Explanation:

As given that water and ammonia both are polar however water is liquid and ammonia is a gas at room temperature.

Let us understand the difference between solid, liquid and gas.

The three states differ from each other due to

a) inter-molecular forces of attraction (the force of attraction between two molecules of same or different substance).

b) the thermal energy of molecules.

In case of solids the inter-molecular forces are maximum and the thermal energy is least. So the molecules are held strong with each other and does not move here and there.

In case of liquid and gas, the inter-molecular forces of attraction are stronger in liquid as compared to gas while thermal energy of gas molecules is more.

In water, the oxygen atom is more electro-negative than nitrogen of ammonia. This makes water molecular more polar.

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The monatomic ions of Groups 1A(1) and 7A(17) are all singly charged. In what major way do they differ? Why?
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<u>Explanation:</u>

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If an atom looses electrons, it leads to the formation of positive ions known as cations. <u>For Example:</u> Sodium is a Group 1 element which looses 1 electron to form Na^+ ions.

Hence, group 1 ions are known as cations and Group 17 ions are known as anions.

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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
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<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
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Answer : The specific heat of tin is, 0.213 J/g.K

Explanation :

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q=m\times c\times (T_{final}-T_{initial})

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c = specific heat capacity of tin = ?

m = mass of tin = 25.0 g

T_{final} = final temperature = 24.5^oC=273+24.5=297.5K

T_{initial} = initial temperature = 99.5^oC=273+99.5=372.5K

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

-399.4J=25.0g\times c\times (297.5-372.5)K

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Therefore, the specific heat of tin is, 0.213 J/g.K

7 0
2 years ago
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