Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get
Now we have to calculate the half-life.
Therefore, the half life of 28-Mg in hours is, 6.94
Correct Answer: Option C:<span> The equilibrium position will shift to the right toward the products.
Reason:
1) This problem is based on </span>Le Chatelier's principle. It is stated as '<em>any</em><span><em> changes in the temperature, volume, or concentration of a system will result in predictable and opposing changes in the system in order minimize this change and achieve a new equilibrium state.</em>'
2) In present case, the reaction involved is:
</span><span> CH3CO2H(aq) + H2O(l) ⇄ CH3CO2-(aq) + H3O+(l)
</span>Hence, when the concentration of acetic acid (reactant) is increased, the equilibrium will shift to right to minimize the effect of change in concentration of reactant.
Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.
¿ cuál es la pregunta que intentas hacer?
