For the reaction N2(g)+O2(g)⇌2NO(g) classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the reaction. Drag the appropriate items to their respective bins.

The question is incomplete, complete question is;

For the reaction:

classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the reaction.

a) half oxygen

b) double oxygen

c) double nitrogen

d) half nitrogen

e) double nitrogen monoxide

f) half nitrogen monoxide

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On an addition of product

a) Half oxygen

Here reactant is reducing, which means that concentration of product is increasing at equilibrium. So, to in order to counter the change the equilibrium will move backward that is leftward direction .

b) Double oxygen

Here reactant is increasing , which means that concentration of product is deceasing at equilibrium. So, to in order to counter the change the equilibrium will move forward that is rightward direction.

c) Double nitrogen

Here reactant is increasing , which means that concentration of product is decreasing at equilibrium. So, to in order to counter the change the equilibrium will move forward that is rightward direction.

d) Half nitrogen

Here reactant is reducing, which means that concentration of product is increasing at equilibrium. So, to in order to counter the change the equilibrium will move backward that is leftward direction .

e) Double nitrogen monoxide

Here product is increasing , which means that concentration of reactant is decreasing at equilibrium. So, in order to counter the change the equilibrium will move backward that is leftward direction.

f) Half nitrogen monoxide

Here product is decreasing , which means that concentration of reactant is increasing at equilibrium. So, in order to counter the change the equilibrium will move forward that is rightward direction.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying