Answer:
Climate, atmosphere, and land
Explanation:
Some of the data collected include air chemistry, temperature, precipitation, cloud cover, and wind speed. Instruments carried on balloons and wind profiling radar provide observations from the surface to more than 10 miles high.
Again great job! They all look correct except 20. is 3.7 due to sig fig of least precision, which you have a mark by!! You don't even need help!;) Here comes the next chemical engineer! :)
Answer:
C is the only reasonable answer.. but this is 6th grade science and I'm in 7th, so I'm pretty sure I'm right
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
Answer: C, 146
Explanation:
NaCl has a molar mass of 58.44 g/mol.
To change moles to g, we do:
2.50 moles × 58.44 g/mol = 146.1 g, so C
