Question:
The options are;
a. Temperature
b. Thermal Energy
c. Hotness
d. Fire Energy
Answer:
The correct option is;
b. Thermal energy
Explanation:
A burner on a stove produces thermal energy which is used to raise the temperature of the metal container (kettle, pot or pans) in which items are placed for heating.
Thermal energy is the internal energy of the system given off as heat which when transferred from one body to another causes the temperature of the receiving body to rise. Thermal energy in a burner is given off when the gaseous fuel reacts or burns in the presence of or with oxygen to produce carbon dioxide and water vapor in an exothermic reaction.
4C + 5H₂ + 13/2O₂ (-125 kJ) → C₄H₁₀ + O₂ → CO₂ + H₂O (-2877 kJ).
Answer:
Explanation:
To answer this question successfully, we need to remember that atoms are neutral species, since the number of protons, the positively charged particles, is equal to the number of electrons, the negatively charged particles. That said, we may firstly find an atom which has 3 electrons (and, as a result, 3 protons, as it should be neutral).
The number of protons is equal to the atomic number of an element. We firstly may have an atom with 3 protons and 3 electrons (atomic number of 3, this is Li).
Similarly, we may take the atomic number of 4, beryllium, and remove 1 electron from it. Upon removing an electron, it would become beryllium cation, 
.
We may use the same logic going forward and taking the atomic number of 5. This is boron. In this case, we need to remove 2 electrons to have a total of 3 electrons. Removal of 2 electrons would yield a +2-charged cation: 
.
i think the answer is C3H8O
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
