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4 years ago

If the carbon atom of hydrocarbon is bonded to less than 4 other atoms, the hydrocarbon is considered _______.

Chemistry

2 answers:

Paul [167]4 years ago

7 0

Answer : The correct option is, (D) unsaturated

Explanation :

As the organic compounds are classified into two categories which are acyclic (aliphatic) and cyclic (aromatic) compounds.

Acyclic compounds are also classified into two categories which are saturated and unsaturated compounds.

Saturated hydrocarbon : These are the compounds which containing single bonds between the carbon-carbon atoms. In saturated compounds, there are more number of hydrogen bonded with the carbon atom.

Unsaturated hydrocarbon : These are the compounds which containing double and triple bonds between the carbon-carbon atoms. In unsaturated compounds, there are less number of hydrogen bonded with the carbon atom.

Aromatic : Aromatic compounds are the type of an unsaturated compounds but it is not necessary that the unsaturated compounds are aromatic. It is a compound which containing the $\pi$ electrons and all the $\pi$ electrons are in conjugation.

Hence, if the carbon atom of hydrocarbon is bonded to less than 4 other atoms, the hydrocarbon is considered as unsaturated.

antiseptic1488 [7]4 years ago

5 0

It is called as unsaturated hydrocarbons

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Answer:

Explanation:

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3 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

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At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i

olganol [36]

Answer:

$K_c=0.0867$

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

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

$Molarity=\frac{0.760}{1.50\ L}$

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

$\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}$