Combustion can be defined as the reaction of a compound with oxygen. The enthalpy of combustion of octane is 
for 
.
<h3>What is the enthalpy of reaction?</h3>
The enthalpy of reaction is the amount of heat energy absorbed or lost by the molecules in the chemical reaction.
The enthalpy of combustion is the amount of heat energy released by the compound in the reaction with oxygen.
The reaction in which heat is liberated with the reaction of a compound with oxygen has an enthalpy of combustion, equivalent to the enthalpy of reaction.
The combustion of octane can be given as:
Thus, the reaction has combustion energy equivalent to the enthalpy of the reaction is . Thus, option B is correct.
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The paths in which electrons travel are called orbitals.
Answer:
A. Thin
Explanation:
i took the test a while back yw <3
According to the calculation, set I is both the most accurate and most precise.
Exactness is a quality or state of being precise. 2a: comparison accuracy sense: the degree of precision with which an action is carried out or a measurement expressed 2b. The novel was fact-checked for historical authenticity. 2a: conformance to truth, to a standard, or to a model: exactness It is impossible to estimate the number of casualties with accuracy. Let's calculate the precision for the fourth set, considering
Δ1 = ∣(8.41−8.56)∣ g=0.15 g
Δ2 = (8.72−8.56) g=0.16 g
Δ3 = ∣(8.55−8.56)∣ g=0.01 g
Therefore: 
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Answer:
108.43 grams KNO₃
Explanation:
To solve this problem we use the formula:
Where
- ΔT is the temperature difference (14.5 K)
- Kf is the cryoscopic constant (1.86 K·m⁻¹)
- b is the molality of the solution (moles KNO₃ per kg of water)
- and<em> i</em> is the van't Hoff factor (2 for KNO₃)
We <u>solve for b</u>:
- 14.5 K = 1.86 K·m⁻¹ * b * 2
Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:
- 275 mL water ≅ 275 g water
- moles KNO₃ = molality * kg water = 3.90 * 0.275
- moles KNO₃ = 1.0725 moles KNO₃
Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:
- 1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃
