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ddd [48]
3 years ago
14

Which forward reaction shows an increase in entropy (disorder)?explain

Chemistry
1 answer:
Lina20 [59]3 years ago
7 0

Explanation:

Entropy also increases when solid reactants form liquid products. Entropy increases when a substance is broken up into multiple parts. The process of dissolving increases entropy because the solute particles become separated from one another when a solution is formed. Entropy increases as temperature increases. S = entropy

k_{b} = Boltzmann constant

\ln = natural logarithm

\Omega = number of microscopic configurations

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Mg (s) + 2 HCl (aq) --> MgCl2 (aq) + H2 (g)
Lera25 [3.4K]

Answer:

m = 190. g MgCl2

Explanation:

Use a molar ratio to get the # of moles MgCl2 produced from 4.00 mol of HCl:

4.00 mol HCl × (1 mol MgCl2/2 mol HCl) = 2.00 mil MgCl2

So 2.00 moles of MgCl2 will be produced. To find the mass in grams, use the molar mass of MgCl2:

2.00 mol MgCl2 × (95.211 g MgCl2/1 mol MgCl2)

= 190. g MgCl2

8 0
3 years ago
If more of a liquid evaporates,does that mean it has stronger intermolecular forces
Gwar [14]

Answer:

No

Explanation:

The stronger the intermolecular forces the molecules tend to stay in the liquid phase. Compounds with weak intermolecular forces are more volatile and more molecules overcome those forces to pass to the gas phase.

7 0
3 years ago
A chlorine Cl and bromine Br atom are adsorbed on a small patch of surface (see sketch at right). This patch is known to contain
MrRissso [65]

Explanation:

It is given that possible number of ways the Cl and Br can be absorbed initially are 100.

S, possible number of ways by which Br can be desorbed is as follows.

                  100 \times 99

Now, we will calculate the change in entropy as follows.

               \Delta S = k_{B} ln (\frac{W}{W_{o}})

where,   k_{B} = Boltzmann constant = 1.38 \times 10^{-23}

             \Delta S = change in entropy

Therefore, we will calculate the change in entropy as follows.

             \Delta S = k_{B} ln (\frac{W}{W_{o}})

                         = 1.38 \times 10^{-23} J/K \times ln (\frac{100}{100 \times 99})

                        = 1.38 \times 10^{-23} J/K \times -4.595

                        = -6.34 \times 10^{-23} J/K

Thus, we can conclude that the change in entropy is -6.34 \times 10^{-23} J/K.

5 0
3 years ago
Help please!!!!
emmasim [6.3K]
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4 0
3 years ago
What kind of graph would best show the info in the data table below?
Rasek [7]

Answer:a line or a circle but i think circle

Explanation:

6 0
3 years ago
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