Answer:
The behavior of molecules in different phases of matter represents a balance between the kinetic energies of the molecules and the attractive forces between them. All molecules are attracted to each other. The molecules are in the solid-state. At higher temperatures, the kinetic energy of the molecules is higher.
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
Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?
Solving:
Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
If: ΔT (T final - T initial) = ?
Answer:
D.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.
Explanation:
For the reaction:
3H₂(g) + N₂(g) → 2NH₃(g)
The enthalpy change is ΔH = -92kJ
This enthalpy change is defined as the enthalpy of products - the enthalpy of reactants. As the enthalpy is <0, The enthalpy of products is <em>lower </em>than the enthalpy of reactants.
Also, it is possible to obtain the enthalpy change from the bond energies of products - bond energies of reactants, thus, The total bond energies of products are <em>lower</em> than the total bond energies of reactants.
The rate of the reaction couldn't be determined using ΔH.
As the bond energy of ammonia is lower than bonds of nitrogen and hydrogen, <em>D. Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.</em>
I hope it helps!
It should be cell, neutron, atom, and electron. If not then I’m so sorry
