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3 years ago

What is the role of the activated complex in a chemical reaction?

Chemistry

1 answer:

brilliants [131]3 years ago

8 0

The activated complex is the intermediate between reactants and products in a chemical reaction

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0

3 years ago

Indicate which of the following will elements will share the electrons and which one will transfer them?

DaniilM [7]

Answer:

Explanation:

8 0

3 years ago

Name the following compound:<br><br> CH3 – CH2 – CH2 – CH2 – CH2 – CH2 – CH2 – CH3

Korvikt [17]

The compound is Octane since it has 8 carbons on the parent chain and has only single carbon to carbon bonds.

7 0

3 years ago

The conversion of glucose to lactic acid is called:

weqwewe [10]

Answer:

Anaerobic glycolysis

Explanation:

is the transformation of glucose to lactate when limited amounts of oxygen (O2) are available.

7 0

2 years ago

Read 2 more answers

Which has the most gravitational potential energy- a 10 kg ball that is 2 meters high or a 10 kg ball that is 4 meters high?

Alenkasestr [34]

Answer:

The 10 kg ball that is 4 meters high is greater than a 10 kg ball that is 2 meters high

Explanation:

From the question,

Gravitational Potential Energy of the first ball

P.E1 = mgh.................... Equation 1

Where P.E1 = Gravitational Potential Energy of the first ball, m = mass of the first ball, h = height of the first ball, g = acceleration due to gravity.

Given: m = 10 kg, h = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E1 = 10×2×9.8

P.E1 = 196 J.

Similarly,

Gravitational potential energy of the second ball is

P.E2 = m'gh'................... Equation 2

Where P.E2 = Gravitational potential Energy of the second ball, m' = mass of the second ball, h' = height of the second ball.

Given; m' = 10 kg, h' = 4 m

Substitute these values into equation 2

P.E2 = 10×4×9.8

P.E2 = 392 J

From the above calculation,

P.E2 is greater than P.E1.

Hence, the 10 kg ball that is 4 meters high is greater than a 10 kg ball that is 2 meters high

3 0

3 years ago