Answer:NASA's Lunar Atmosphere and Dust Environment probe, for instance, orbits the moon and collects information about its atmosphere and surface. This important data can help scientists learn more about moons, asteroids and other objects in the solar system.
Explanation:
Answer: How many grams are in 2.5 moles of N2?
Explanation: 1 mole is equal to 1 moles N2, or 28.0134 grams.
One mole of N2 molecules would have a mass of 2 X 14.01 g = 28.02 g.
When q is the heat energy in joules (J)
so, according to this formula, we can get q (in joule unit):
q = M*C*ΔT
when M is the mass of the water sample = 1.85 g
C is the specific heat capacity of water = 4.18 J/g.°C
and Δ T is the difference in temperature (Tf-Ti) = 33 - 22 = 11°C
So, by substitution, we will get the value of q ( in Joule):
∴ q = 1.85 g * 4.18 J/g.°C * 11 °C
= 85 J
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
It would be the water based carbon cycle
