Answer:
The answer is B. Van der Waals forces are weaker than ionic and covalent bonds.
Explanation:
In general, if we arrange these molecular forces from the strongest to weakest, it would be like this:
Covalent bonds > Ionic bonds > Hydrogen bonds > Dipole-Dipole Interactions > Van der Waals forces
Covalent bonds are known to have the strongest and most stable bonds since they go deep and into the inter-molecular state. A diamond is an example of a compound with this characteristic bond.
Ionic bonds are the next strongest molecular bond following covalent bonds. This is due to the protons and electrons causing an electro-static force which results to the strong bonds. An example would be Sodium Chloride (NaCl), which when separated is Na⁺ and Cl⁻.
Van der Waals forces, also known as Dispersion forces, are the weakest type of molecular bonds. They are only formed through residual molecular attractions when molecules pass by each other. It doesn't even last long due to the uneven electron dispersion. It can be made stronger by adding more electrons in the molecule. This kind of molecular bonds appear in non-polar molecules such as carbon dioxide.
HOPE THIS HELPS!!!!!!!!!!!!!!
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The paths in which electrons travel are called orbitals.
Answer:
Chloroform= limiting reactant
0.209mol of CCl4 is formed
And 32.186g of CCl4 is formed
Explanation:
The equation of reaction
CHCl3 + Cl2= CCl4 + HCl
From the equation 1 mol of
CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4
From the question
25g of CHCl3 really with Cl2
Molar mass of CHCl3= 119.5
Molar mass of Cl2 = 71
Hence moles of CHCl3= 25/119.5 = 0.209mol
Moles of Cl2 = 25/71 = 0.352mol
Hence CHCl3 is the limiting reactant
Since 1 mole of CHCl3 gave 1mol of CCl4
It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4
Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g
This question is describing the following chemical reaction at equilibrium:
And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:
Thus, by recalling the Van't Hoff's equation, we can write:
Hence, we solve for the enthalpy change as follows:
Finally, we plug in the numbers to obtain:
![\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-8.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%2Aln%280.25%2F9%29%7D%7B%5B%5Cfrac%7B1%7D%7B%2875%2B273.15%29K%7D%20-%5Cfrac%7B1%7D%7B%2825%2B273.15%29K%7D%20%5D%20%7D%20%5C%5C%5C%5C%5C%5C%5CDelta%20H%3D4%2C785.1%5Cfrac%7BJ%7D%7Bmol%7D)
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Answer:
Distance = 200 km
Distance = 204 km
Speed = 77 km/h
Time = 21.42 h
Explanation:
Given:
A.
Speed = 100 km/h , Time = 2 h
Find:
Distance
B.
Speed = 68 km/h , Time = 3 h
Find:
Distance
C.
Distance = 154 km , Time = 2 h
Find:
Speed
D.
Distance = 1500 km speed = 70 km/h
Find:
Time
Computation:
Speed = distance / time
A.
Distance = 100 x 2
Distance = 200 km
B.
Distance = 68 x 3
Distance = 204 km
C.
Speed = 154 / 2
Speed = 77 km/h
D.
Time = 1500 / 70
Time = 21.42 h
