It’s ionic
KCl(at)
Potassium chloride
A. phase changing. Physical change is any change in matter that involves the substance going from one physical state to another, and phase change is most commonly used to describe transitions between solid, liquid and gaseous states of matter.
The shells further away from the nucleus are LARGER and can hold MORE electrons
In general, we have this rate law express.:
![\mathrm{Rate} = k \cdot [A]^x [B]^y](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5Ey)
we need to find x and y
ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).
then we go to compare two experiments in which only one concentration is changed
compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]
Δ[B]= 0.3 / 0.1 = 3
now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:
ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...
solve for y in the equation
![\Delta \mathrm{Rate} = \Delta [B]^y](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BB%5D%5Ey)

To this point,
![\mathrm{Rate} = k \cdot [A]^x [B]^1](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5E1%20)
do the same to find x.
choose two experiments in which only the concentration of B is unchanged:
Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4
solve for x for
![\Delta \mathrm{Rate} = \Delta [A]^x](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BA%5D%5Ex)

the rate law is
Rate = k·[A]²[B]
Answer:
Increasing the temperature
Explanation:
According to Le Chatelier's principle, "if any of the conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".
Since the forward reaction is exothermic, this suggests that heat is given off in the reaction.
Increasing the temperature will favors the reverse process which is endothermic.
In going from the reactants to the the products, the reaction is deemed exothermic which suggests that it produces heat.
The backward reaction is endothermic i.e it absorbs heat. If the temperature is increased, this process will be favored and more reactants produced. Less of the diiodine pentoxide will form. There is a leftward shift of the equilibrium