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3 years ago

There are σ bonds and π bonds in h3c-ch2-ch=ch-ch2-c≡ch.

Chemistry

1 answer:

FrozenT [24]3 years ago

6 0

The number of sigma and pi bonds are,

Sigma Bonds = 16

Pi Bonds = 3

Explanation:
Every first bond formed between two atoms is sigma. Pi bond is formed when already a sigma bond is there. While in case of Alkyne (triple Bond) there is one sigma and one pi bond already present, so the third bond is formed by second side-to-side overlap of orbitals, hence, a second pi bond is formed.
Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.

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Which answer choice lists the number and type of elements in a compound in the way that is close to their actual arrangement?

alexandr1967 [171]

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3 years ago

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3 years ago

Calculate the molarity of sodium ion in a solution made

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Answer:

0.1035 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Sodium chloride will furnish Sodium ions as:

$NaCl\rightarrow Na^{+}+Cl^-$

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

$Moles =0.288 \times {3.58\times 10^{-3}}\ moles$

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

$Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}$

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

$Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles$

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}$

<u> The final concentration of sodium anion = 0.1035 M</u>

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3 years ago

There are ________ σ bonds and ________ π bonds in h3c-ch2-ch=ch-ch2-c≡ch.

There are σ bonds and π bonds in h3c-ch2-ch=ch-ch2-c≡ch.