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luda_lava [24]
3 years ago
11

For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.

Chemistry
1 answer:
vichka [17]3 years ago
6 0

Answer:

\Delta G =-103.95kJ

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

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

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ

Best regards!

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A solution is prepared by dissolving 15 grams of sugar in 85 grams of water. How many percent of sugar is in the solution?
trasher [3.6K]

Answer:

15%

Explanation:

From the question given above, the following data were obtained:

Mass of sugar = 15 g

Mass of water = 85 g

Percentage of sugar in the solution =?

Next, we shall determine the mass of the solution. This can be obtained as follow:

Mass of solute (sugar) = 15 g

Mass of solvent (water) = 85 g

Mass of solution =?

Mass of solution = mass of solute + mass of solvent

Mass of solution = 15 + 85

Mass of solution = 100 g

Finally, we shall determine the percentage of the sugar in the solution. This can be obtained as follow:

Mass of solute (sugar) = 15 g

Mass of solution = 100 g

Percentage of sugar in the solution =?

Percentage = solute /solution × 100

Percentage = 15 / 100 × 100

Percentage of sugar in the solution = 15%

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3 years ago
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Photosynthesis converts solar radiation from the sun to chemical energy in green plants. Another group of organisms in the ecosystem, the primary consumers, feed directly on the producers and as such, are able to gain some energy from them.

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Answer:

Step 1

1 of 2

V = m / D; V (nugget of gold) = 50 g / 5.0 g / cm 3 = 10 cm3; V (nugget of iron pyrite) = 50 g / 19 g / cm3 = 2.63 cm 3

Result

2 of 2

Due to these results you can conclude that nugget of gold which volume is 10 cm3 is bigger than volume of nugget of iron pyrite which is 2.63 cm3.

Explanation:

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svlad2 [7]

Answer:

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Explanation:

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How do properties of matter help us choose the best material for the job? Use examples.
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