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3 years ago

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Which sequence of atomic numbers represents elements that have similar chemical proprieties

1 answer:

Kazeer [188]3 years ago

5 0

Hey. Hey. You know there’s something called google?
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Stolb23 [73]

Inherited it from his grandfather

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3 years ago

Is a chemical reaction when put salt into fizzy drinks and why

fomenos

Um to my best knowledge I believe it is the carbonated water

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3 years ago

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Why do you think some of the indicators used in experiment 2 were different than the ones used in experiment 1?

pentagon [3]

During selection of indicator. We choose an indicator which have pH range equivalent to the pH change of reaction to give better result and better observation.

So there are some different indicator are used in table 2 as compared to the table 1.

- Alizarin and phenolphthalein are basic indicator and their pH range is more than 8 so they are used in table 2

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3 years ago

Which characteristic is not used to classify stars? size color distance from Earth age

Westkost [7]

According to the topic on stars, galaxies and the universe, the characteristics that are used to classify stars are color, temperature, size, elemental or compound composition, and brightness. Among these factors give, the characteristic not used to classify stars is distance from the earth.

8 0

3 years ago

Read 2 more answers

In a unimolecular reaction with twice as much starting material as product at equilibrium, what is the value of Keq? Is ΔG o pos

lubasha [3.4K]

Answer : The value of $K_{eq}$ is, 0.5 and $\Delta G$ = positive.

Explanation :

The unimolecular reaction is:

$A\rightarrow B$

In unimolecular reaction, the starting material is 2 times to the product.

$A=2B$ .........(1)

As we know that:

$K_{eq}=\frac{B}{A}$ ...........(2)

Now substitute equation 1 in 2, we get:

$K_{eq}=\frac{\frac{A}{2}}{A}$

$K_{eq}=0.5$

Now we have to calculate the value of $\Delta G^o$ at 298 K.

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = 0.5

Now put all the given values on the above formula, we get:

$\Delta G^o=-(8.314J/mol.K)\times (298K)\ln (0.5)$

$\Delta G^o=1717.32J/mol$

Thus, the value of $\Delta G^o$ at 298 K is, 1717.32 J/mol

As we know that:

$\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

Thus, the $\Delta G$ = +ve. So, the reaction is non spontaneous.

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4 years ago