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3 years ago

Which of the following would most likely happen if water did not form hydrogen bonds?

Chemistry

2 answers:

Oksanka [162]3 years ago

8 0

Water would not expand when it freezes.

stellarik [79]3 years ago

3 0

The hydrogen bond is a secondary bond formed between a hydrogen (attached to an highly electronegative element ,: F, O or N) and another atom of an highly electronegative element ,: F, O or N.

In case of water hydrogen bond helps water

a) to dissolve certain substances which can make hydrogen bond

b) it makes it liquid and room temperature unlike H2S which is a gas at room temperature due to absence of hydrogen bond

c) higher volume and low density of ice as compared to liquid water. Due to hydrogen bond ice forms open cage like structure which increases its volume and decreases its density. Thus water expands on freezing

So in absence of hydrogen bond Water would not expand when it freezes.

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I need your help!! about Chemistry (15 point!!)

irakobra [83]

1. I think it is true?
2. Low melting points
3. True
4. Atomic number, I think it’s periods?
5. Groups?

Sorry, I might not get all of them right :(

Hope this helps you in any way!!

6 0

3 years ago

. 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]

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

Explanation:

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

Element A has 15 protons. Determine the number of neutrons in its isotope with mass number 33.

bagirrra123 [75]

In an electrically neuteral atom, number of protons = number of electrons = atomic number.
Mass number = neutrons + protons/electrons/atomic number
Therefore,
neutrons = mass number - protons/electrons/atomic number
Neutrons = 33 - 15 = 18

The answer is thus B. But this is the solution and explanation along with it as proof.

6 0

3 years ago

Look at the periodic table. Find the element carbon. What is its symbol?

Sliva [168]

A. ca is the answer because its short for carbon

6 0

4 years ago

Read 2 more answers

Which has a smaller atomic radius: antimony (Sb) or bismuth (Bi)?

son4ous [18]

Bi

The nitrogen family includes the following compounds: nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi).

7 0

3 years ago

Read 2 more answers