Movement from place to palce
The correct answer from the choices given is the last option. The can from the <span> car will lose the carbon more quickly because there are fewer solute–solvent collisions. The can in the car has a lower temperature than the one in the refrigerator. At low temperature, the solubility of carbon dioxide in the liquid decrease therefore particles would tend to be in the vapor phase and escape from the liquid.</span>
Answer:
The reaction will move to the left.
Explanation:
<em>Ba(OH)₂ = Ba²⁺ + 2OH⁻,</em>
<em>Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.</em>
- If H⁺ ions are added to the equilibrium:
H⁺ will combine with OH⁻ to form water.
<em>So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.</em>
<em />
<em>According to Le Châtelier's principle: </em>when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
- So, the reaction will move to the right to suppress the effect of decreasing OH⁻ concentration.
- The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.
<em>So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.</em>
This is false. One mole of a gas occupies 22.4 L at STP, which is taken to be 0°C (273 K) and 1 atm. If atmospheric conditions depart from these values, this assumption cannot be used.
Answer: 25g/ml
Explanation:
Density = 100.0g/ 4.0mL = 25g/mL
