A molecule of an unsaturated hydrocarbon must have(1) at least on single carbon-carbon bond

Chemistry

2 answers:

tiny-mole [99]2 years ago

8 0

Answer:

(2) at least one multiple carbon-carbon bond

Explanation:

Hydrocarbons are organic compounds containing carbon (C) and hydrogen (H) atoms. Here, the skeletal structure is defined by the carbon-carbon bonds in which the C atoms can be linked to form linear chains or cyclic structures.

The C-C bonds in hydrocarbons are covalent in nature. The bonds can be single or multiple bonds. Structures in which the C atoms are linked by single bonds are called saturated hydrocarbons whereas multiple bonds between C atoms results in unsaturated hydrocarbons.

Therefore, a molecule of an unsaturated hydrocarbon must have at least one carbon-carbon multiple bond. The simplest example is ethene (C2H4) in which the C-C atoms are connected by a double bond.

Ethene → H2C=CH2

Romashka-Z-Leto [24]2 years ago

6 0

An unsaturated hydrocarbon has a double or triple C-C bond as unsaturated literally means it is not saturated with atoms so there are "spare bonds" left.

Therefore:

A molecule of an unsaturated hydrocarbon must have 2 - at least one multiple Carbon-Carbon bond.

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levacccp [35]

Answer:

Explanation:

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8 0

2 years ago

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II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo

Leto [7]

Answer:

$Excess=3.53g$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2Mg(s)+O_2(g) \rightarrow 2MgO(s)$

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

$n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO$

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

$m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg$