Answer:
A. 2C + H₂ ⟶ CH₄
Explanation:
A. 2C + H₂ ⟶ CH₄
UNBALANCED. 2C on the left and 1C on the right
B. 2Al₂O₃ ⟶ 4Al + 3O₂
Balanced. Same number of each type of atom on each side.
C. 2H₂O₂ ⟶ 2H₂O + O₂
Balanced. Same number of each type of atom on each side.
D. 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O
Balanced. Same number of each type of atom on each side.
Explanation:
atom changes from a ground state to an excited state by taking on energy from its surroundings in a process called absorption. The electron absorbs the energy and jumps to a higher energy level. In the reverse process, emission, the electron returns to the ground state by releasing the extra energy it absorbed
All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.<span>
Freezing point depression or Boiling point elevation:
</span><span>ΔT = -K (m) (i)
</span>ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent.
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K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent.
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m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent.
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i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.</span>