Two changes would make this reaction reactant-favored
C. Increasing the temperature
D. Reducing the pressure
<h3>Further explanation</h3>
Given
Reaction
2H₂ + O₂ ⇒ 2H₂0 + energy
Required
Two changes would make this reaction reactant-favored
Solution
The formation of H₂O is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)
And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater
Answer:9.49g/mL
Explanation:
Mass of toy = 43.672g
Volume of water = 34.4mL
Volume of toy + volume of water = 39mL
Volume of toy = 39 — 34.4 = 4.6mL
Density = Mass /volume
Density = 43.672/4.6 = 9.49g/mL
Explanation:
The speed of molecules increases when temperature is increased as it will result in more number of collisions between the molecules. Thus, there will be increase in kinetic energy of molecules and increase in the speed of solvent molecules.
Whereas on decreasing the temperature, the kinetic energy of molecules will decrease. This will result in less number of collisions between the molecules. Therefore, the speed of solvent molecules will slow down.
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