Select the molecule where free rotation around one or more bonds is not possible. Group of answer choices N2H4 C2Cl4 NH3 C2H6 CC
1 answer:
Answer:
C₂Cl₄
Explanation:
To know if free rotation around a bond in a compound is possible, we need to see the structure of the compound (picture in attachment).
In single bonds, which are formed by σ bonds, the atoms are not fixed in a single position, and free rotation is permitted.
Double and triple bonds are formed by a σ bond and one or two π bonds, respectively. These bonds do not allow rotation, since it is not possible to twist the ends without breaking the π bond.
The chloroethylene (C₂Cl₄) has two carbons with an sp2-sp2 hybridization, they are bonded together by a double bond. <u>Free rotation on this bond is not possible, because six atoms, including the carbon atoms, doubly bonded and the four chlorine atoms bonded to them, must be on the same plane. </u>
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Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Mg²⁺: 1 mole
- Na₂CO₃: 1 mole
- MgCO₃: 1 mole
- Na⁺: 2 moles
The molar mass of the compounds is:
- Mg²⁺: 24.3 g/mole
- Na₂CO₃: 106 g/mole
- MgCO₃: 84.3 g/mole
- Na⁺: 23 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
- Na₂CO₃: 1 mole ×106 g/mole= 106 grams
- MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
- Na⁺: 2 moles ×23 g/mole= 46 grams
The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?
<u><em>mass of magnesium= 2.13 grams</em></u>
Finally, 2.13 grams of magnesium was dissolved in the solution.
Learn more about the reaction stoichiometry: