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zheka24 [161]
2 years ago
14

Which compound contains both sigma and pi bonds? HCCl3 H2CO H2S HBr.

Chemistry
1 answer:
Aliun [14]2 years ago
6 0

The compound that contains both sigma and pi bonds has been \rm \bold{H_2CO}. Thus, option B is correct.

The compounds have been resulted by the sharing of the valence electrons between atoms, for the completion of octet of each elements.

The bonds can be saturated with the presence of only sigma bond. The unsaturated bonds has presence of pi bonds as well. The bond with one pi and one sigma has been a double bond, while 1 sigma and 2 pi has been a triple bond.

The bonds present in the following structures has been:

\rm HCCl_3=4\;\sigma\;bonds\\H_2CO=2\;\sigma,\;1\;\pi\;bond\\H_2S=2\;\sigma\;bonds\\HBr\;=1\;\sigma\;bond

The compound with the presence of both sigma and pi bonds has been \rm \bold{H_2CO}. Thus, option B is correct.

For more information about the sigma and pi bonds, refer to the link:

brainly.com/question/14018074

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Lilit [14]

Answer:

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Explanation:

Divide the mass by the volume in order to get an object's Density.

so, 23.2 divided by 1.20 mm3

=1.93333333 g/cm3

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4 0
3 years ago
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Serga [27]

Answer:

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Explanation:

Hello,

Standard enthalpy of reaction is defined by the difference between the products whole enthalpy and the reagents whole enthalpy. It is known that an element's (by itself or as a diatomic gas) enthalpy of formation is 0, turning out into an equality between the enthalpy of formation and the products whole enthapy. Thus, the reactions that accomplish the given questions are 1, 2, 3 and 4, since the reagents are just the elements sodium, chlorine, carbon and oxygen.

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Complete the following road map for converting volume of A to volume of B for a titration of aqueous solution A with aqueous sol
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Answer:

Explanation:

The solution of known concentration is expressed as molarity. Molarity is the mole fraction of solute (i.e. the dissolved substance) per liter of the solution, Molarity is also commonly called molar concentration.

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\mathtt{Molarity = \dfrac{moles \ of \ solute}{ liters \ of \ solution}}

To copy and complete the road map from the given question, we have the following array:

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d. multiplied by the molarity of A

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          moles A

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b. multiplied by the moles of B / moles of A

               ↓

         moles B

               ↓

c. divided by the molarity of B

               ↓

        volume B (L)

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4 years ago
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