Answer: Thus 0.724 mol of
are needed to obtain 18.6 g of 
Explanation:
To calculate the moles :

According to stoichiometry :
2 moles of
are produced by = 1 mole of 
Thus 1.09 moles of
will be produced by =
of 
But as yield of reaction is 75.6 %, the amount of
needed is =
Thus 0.724 mol of
are needed to obtain 18.6 g of 
D - density: 13,534 g/ml
m - mass: 10g
V - volume: ??
_____________
d = m/V
V = m/d
V = 10/13,534
V = 0,7389 ml
:•)
Answer: i think its protons and electrons but it also might just be atoms because protons and electrons make atoms when there are also neutrons
Explanation:
Answer:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Explanation:
Hello,
In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

The suitable equilibrium constant turns out:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Or in terms of the initial equilibrium constant:

Since the second reaction is a doubled version of the first one.
Best regards.