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4 years ago

What is the general formula for the homologous series that includes ethene?

Chemistry

1 answer:

Masja [62]4 years ago

8 0

Answer:

See below

Explanation:

The common formula of the same alkene series is CnH2n, where n is the quantity of atoms of carbon. Owing to the hydrocarbons of alkenes with at least one double carbon relation, the alkene-like sequence starts with ethen C2H4.

<em><u>Hope this helps.</u></em>

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The pH indicators to be used are Phenolphthalein, Red cabbage, Bromthymol blue and Congo red.

<h3>What are pH indicators?</h3>

Indicators are substances which change color as the pH of a medium changes.

The common indicators and their pH range is as follows:

Phenolphthalein - pH range of 8.3 and 10.5
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Learn more about pH indicators at: brainly.com/question/13779537

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2 years ago

The company's chief financial officer recognizes the need for an upgrade to the smart watches but does not understand why the bu

goldenfox [79]

Answer:

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Explanation:

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3 years ago

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8090 [49]

Answer: True

Explanation:

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3 years ago

H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

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