Answer:
2.256 moles of oxygen atoms
Explanation:
CaCo3 is calcium carbonate
every mole of calc carb has 3 moles of O atoms
.752 moles of this has 3 times as many moles of Oxygen atoms
Answer:
36.2 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 8.6 atm
- Initial temperature of the gas (T₁): 38°C
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final temperature of the gas (T₂): ?
Step 2: Convert T₁ to Kelvin
We will use the following expression.
K = °C +273.15
K = 38 °C +273.15 = 311 K
Step 3: Calculate T₂
We will use Gay Lussac's law.
P₁/T₁ = P₂/T₂
T₂ = P₂ × T₁/P₁
T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K
Answer:
3.48 g NaF
Explanation:
Dimensional Analysis:

Answer (Simplifying the Expression Above):
3.48 g NaF
Answer: +178.3 kJ
Explanation:
The chemical equation follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaO%28s%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5E0f_%7BCO_2%7D%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaCO_3%28s%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-635.1%29%29%2B%281%5Ctimes%20%28-393.5%29%29%5D-%5B%281%5Ctimes%20%28-1206.9%29%29%5D)
The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ