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Answer:</h3>
8CO₂
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Explanation:</h3>
We are given;
- Butane is a hydrocarbon in the homologous series known as alkane.
We are required to determine the other product produced in the combustion of butane apart from water.
- We know that the complete combustion of alkane yields carbon dioxide and water.
- Therefore, combustion of butane will yield carbon dioxide and water.
- The balanced equation for the complete combustion of butane will be;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Answer:
Balanced reaction:
3 H2 (g) + N2 (g) → 2 NH3 (g)
Use stoichiometry to convert g of H2 to g of NH3. The process would be:
g H2 → mol H2 → mol NH3 → g NH3
12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3
Explanation: See above
Hope this helps, friend.
Flammable gases are gases that have the tendency to <u>explode (burst into flames)</u> when they come in contact with the <u>appropriate quantities of air, oxygen, or any suitable oxidant.</u>
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Devices that mix air or oxygen with flammable gasses cannot be used unless approved by an <u>authorized or approved personnel.</u>
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- Examples of flammable gases are hydrocarbons such as <u>Propane, Acetylene</u>, e.t.c.
- Flammable gases are dangerous to humans and the environment and if they are not handled properly, can cause severe damages and even death.
- During the preparation and usage of flammable gases, safety precautions should be properly enforced such as the <u>wearing of appropriate safety gear and goggles</u>.
- Devices that mix air or oxygen with flammable gases cannot be used unless approved by an <u>authorized or approved personnel.</u>
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To learn more, visit the link below:
brainly.com/question/3702349
Answer:
If you increase the mass at a given force the rate of acceleration slows. Therefore, mass is inversely proportional to acceleration.
Explanation:
Thats the answer I need brainliest PLEASE jk.
Answer:
-122 J/K
Explanation:
Let's consider the following balanced reaction.
N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)
We can calculate the standard reaction entropy (ΔS°) using the following expression.
ΔS° = Σ ηp × Sf°p - Σ ηr × Sf°r
where,
- η: stoichiometric coefficients of products and reactants
- Sf°r: entropies of formation of products and reactants
ΔS° = 2 mol × 240.06 J/K.mol - 1 mol × 191.61 J/K.mol - 2 mol × 205.14 J/K.mol
ΔS° = -121.77 J/K ≈ -122 J/K