Explanation:
Due to the positive value of the change in temperature, this is an endothermic reaction.
Since the forward reaction is endothermic, increasing the temperature increases the equilibrium constant (k).
In an equilibrium system, the position of the equilibrium will move in a way to annul the change made to the system. An increase in temperature for an endothermic reaction would favour the reaction, leading to increase in amount of products and decrease in amount of reactants.
Answer:
They also showed the effects of pressure on volume if temperature stayed the same
Explanation:
They also showed the effects of pressure on volume if temperature stayed the same is the experiment that will provide an evidence for Boyle's law.
Boyle's law states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".
- The law is an affirmation of what happens when there is a dynamics between pressure and volume if temperature is made constant.
- So the experiment designed to investigate this proves and shows Boyle's law.
Answer:
52 da
Step-by-step explanation:
Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.
The i<em>ntegrated rate law for a first-order reaction </em>is
ln([A₀]/[A] ) = kt
Data:
[A]₀ = 750 mg
[A] = 68 mg
t_ ½ = 15 da
Step 1. Calculate the value of the rate constant.
t_½ = ln2/k Multiply each side by k
kt_½ = ln2 Divide each side by t_½
k = ln2/t_½
= ln2/15
= 0.0462 da⁻¹
Step 2. Calculate the time
ln(750/68) = 0.0462t
ln11.0 = 0.0462t
2.40 = 0.0462t Divide each side by 0.0462
t = 52 da
Answer:
12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N
Explanation:
The reaction is:
6Li(s) + N₂(g) → 2Li₃N(s)
If nitrogen is in excess, the lithium is the limiting reactant.
Ratio is 2:6
2 moles of nitride were produced by 6 moles of Li
Then, 0.6 moles of nitride were produced by (0.6 .6)/ 2 = 1.8 moles of Li
Let's convert the moles to mass → 1.8 mol . 6.94 g/ 1mol = 12.5 g of Li
