How does the picture show that the inner core spins faster than the outer core?
Answer:
1. light is made up of energy
2. light travels in a straight line
3. the speed of light is exactly 299 792 km per second
4. this is the speed when light is traveling in a vacuum and not obstructed by the atmosphere 5. traveling at the speed of light you could go around the earth seven and a half times in a second
6. light can move super fast super slow and not at all
7. we can use light to weigh Stars
8. we can use light as tweezers
9. bubbles can turn sound into light
10. lasers can make things cold
Answer:
The correct answer is - 5 carbon compounds due to low to high intermolecular forces between their molecules.
Explanation:
Bottle C has gas in it and we know that alkane has carbon and hydrogen only which means they have a single sigma bond between them and very low intermolecular forces in between molecules and are present mostly at gaseous state. Thus, bottle C has alkane.
Alcohols have -OH group that can form rarely two pi bonds which means they have intermediate intermolecular force whereas acids have -cooH group with a high molecular force so bottle B with liquid is alcohol and A has acid.
Fusion occurs constantly on our sun, which produces most of its energy via the nuclear fusion of hydrogen into helium. Neither do fusion reactions produce the large amounts of dangerous radioactive waste that fission reactions do. That's why it's such a dreamy source of energy.
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
