Answer:
Explain: In order to balance the chemical equation, you need to make sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. In order make both sides equal, you will need to multiply the number of atoms in each element until both sides are equal.
Hello,
Your questions states:
During a change of state, the temperature of a substance _____?
In which you gave us some choices:
A. decreases if the arrangement of particles in the substance changes.
B. remains constant until the change of state is complete.
C. increases if the kinetic energy of the particles in the substance increases.
D. increases during melting and vaporization and decreases during freezing and condensation.
Your answer would be:
B. remains constant until the change of state is complete.
Your explanation/Reasoning:
It absorbs the energy, then after the phase changes it then increases the temperature all over again.
Have a nice day:)
Hope this helps!
~Rendorforestmusic
Answer:
![\large \boxed{\text{528.7 g} }](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B528.7%20g%7D%20%7D)
Explanation:
It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.
Then you can consider it to be 11018 "moles" of "kJ"
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32.00
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ
n/mol: 7280
1. Moles of O₂
The molar ratio is 25 mol O₂:11 018 kJ
![\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20O%7D_%7B2%7D%20%3D%20%5Ctext%7B7280%20kJ%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B25%20mol%20O%7D_%7B2%7D%7D%7B%5Ctext%7B11%20018%20kJ%7D%7D%20%3D%20%5Ctext%7B16.52%20mol%20O%7D_%7B2%7D)
2. Mass of O₂
![\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20C%24_%7B8%7D%24H%7D_%7B18%7D%20%3D%20%5Ctext%7B16.52%20mol%20O%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B32.00%20g%20O%7D_%7B2%7D%7D%7B%5Ctext%7B1%20mol%20O%7D_%7B2%7D%7D%20%3D%20%5Ctextbf%7B528.6%20g%20O%7D_%7B2%7D%5C%5C%5Ctext%7BThe%20reaction%20requires%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B528.67%20g%20O%7D_%7B2%7D%7D%24%7D)
The person above me is correct I took a test on this so it’s the right answer