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elena55 [62]
3 years ago
11

When you squeeze an air-filled balloon, what happens inside?

Chemistry
1 answer:
Harman [31]3 years ago
4 0

Option D

When you squeeze an air-filled balloon, what happens inside: There are more collisions of air molecules against the wall of the balloon.

<u>Explanation:</u>

If you compress off the balloon, one seemingly sense the air forcing up on the wall of the balloon with indeed more imposing power. This rise in force is due to a drop in quantity. By squeezing the balloon, you lessen the area the gas bits can hold.

As the particles are driven a little closer collectively, they oppose more, so the force from the moving gas bits rises.  Boyle’s Law pronounces that the quantity of a determined quantity of gas limits as its load rises. If the quantity rises, its load reduces.

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What happens when a solid is dissolved in a liquid
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the solid particles take up the intermolecular spaces in the liquid.

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Which one of the following systems has the highest entropy?
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Answer: a

Explanation:

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How to determine whether an element is paramagnetic or diamagnetic?
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To determine whether an element is paramagnetic or diamagnetic, you need to examine the electron configuration of the element. If it has unpaired electrons, then the substance is paramagnetic. If the electrons are paired, then it is diamagnetic.
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3 years ago
What is the molarity of 4 qof NaCl (MM=58.45) in 3,800 mL of solution?
tiny-mole [99]

Answer:

.018 M

Explanation:

grams/MM=ans./volume(L) = M

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8 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]

<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
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