Answer is: ammonia experience only dispersion intermolecular forces with BF₃ (boron trifluoride) because BF₃ is only nonpolar molecule (vectors of dipole moments cansel each other, dipole moment is zero).
The London dispersion force (intermolecular force) <span>is a temporary attractive </span>force between molecules.
Answer:
I know for a fact the correct answer is B
The potassium will donate one of its valence electrons
Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
Answer:
The least substituted product (anti-Markovnikov)
Explanation:
The ROOR is used in the addition reaction of HBr to an organic substance (an alkene for example).
In normal conditions (with no ROOR) the adition of the halogen will be performed in the most substituted C (following the rule of Markovnikov that says that the stability increases with the more substituted is the C).
But in presence of ROOR, the reaction takes other mechanism (free radicals), and the product in this case is the one with the Br added in the least substituted C.
