Answer:
zero order reaction
Explanation:
The seem to be a reaction is zero order as the data only fits to zero order integrated rate law.
Integrated rate law:
[A] = [A]_0 – kt
Here y = [ A] and slope is -k , so when we plot concentration of A with respect to t then we get the equation in the form of line
y = mx + c
So the answer Zero order in [A]
Answer: Options (a) and (d) are the correct answer.
Explanation:
A catalyst is the substance which helps in increasing the rate of reaction.
Activation energy is the minimum amount of energy required by reactants to start the reaction. On addition of catalyst, the path of reaction changes because the energy barrier gap reduces and hence, the activation energy also decreases.
In the absence of catalyst, we need to increase the temperature so that reaction can occur quickly.
Whereas on addition of catalyst, there is no need to increase the temperature as the catalyst itself is sufficient to increase the rate of reaction. As a result, temperature should be lowered when there is addition of catalyst in the reaction.
Thus, we can conclude that catalysts can save money by essentially lowering the activation energy and temperature required.
<span>The ammonia molecules have weaker inter molecular bonds then the hydrogen and nitrogen, this means that the ammonia condensed at a lower temperature so will be a liquid that will be separated from the reaction, while the nitrogen and hydrogen remain as gases so are easy are recycle back into the reacting chamber.</span>
Answer:
The elements that combine to form binary molecular compounds are both nonmetal atoms. This contrasts with ionic compounds, which were formed from a metal ion and a nonmetal ion. ... Prefixes are used in the names of binary molecular compounds to identify the number of atoms of each element.
Answer:
328 ml
Explanation:
We have given final volume =575.2 ml=0.575 L
Final concentration = 0.8012 M
We know that moles of copper(II) nitrate = final volume ×final concentration =0.8012×.0575=0.4606 moles
We have to find initial volume
So initial volume 
