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mixas84 [53]
2 years ago
12

draw the lewis structure, including unshared pairs, of the following molecule. carbon has four bonds in the compound. propane (c

3h8)]
Chemistry
1 answer:
SIZIF [17.4K]2 years ago
6 0

There are three carbon and 8 hydrogen present in propane molecule. The Lewis structure of propane is shown as:

Three molecules of such carbon atoms bound to eight molecules with hydrogen atoms make up the organic complex propane molecule.

It is known that carbon has 4 valence electrons and hydrogen has one valence electron. Carbon needs 4 extra electrons to complete its octet hence, it will share its electrons with with 4 hydrogen atom and complete its octet.

Carbon will be formed 4 bond . Three bond with hydrogen and one bond with carbon atom.

To know more about Lewis structure.

brainly.com/question/15837141

#SPJ4

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Two processes are described below:
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Answer:

i would say D i just did this but i kinda forgot so sorry if im wrong or A

Explanation:

4 0
3 years ago
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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

4 0
3 years ago
A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?
uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

  • Hafnium = 55.7% = 55.7g
  • Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

  • Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
  • Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

  • Hafnium = 0.312 ÷ 0.312 = 1
  • Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

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7 0
1 year ago
Natural atom of the same element may have the same _________?
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3 years ago
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Consider the balanced chemical reaction below and determine the percent yield of sodium bromide if 2.36 moles of iron(iii) bromi
hammer [34]
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the  %  yield  =  actual  yield/  theoretical  yield  x  100

that  is    6.14/7.08  x100=  86.72%
8 0
4 years ago
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