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).
Peter is taller, convert cm to in.
The accuracy is associated with magnitude of fractional error or relative error in that measurement
the precision of a measurement is associated with least count of the measuring instrument.
precision does not mean accuracy but both are highly demanded while making measurements in physics.
Answer: 460.624
Explanation:
1. Multiply the numbers
(24.5260 x 2.56) + 397.84
= (62.784) + 397.84
2. Add the numbers
(62.784) + 397.84
= 460.624
Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.