Answer: 0.294 mol of 
present in the reaction vessel.
Explanation:
Initial moles of 
= 0.682 mole
Initial moles of = 0.440 mole
Volume of container = 2.00 L
Initial concentration of 
Initial concentration of 
equilibrium concentration of 
The given balanced equilibrium reaction is,
Initial conc. 0.341 M 0.220 M 0 M
At eqm. conc. (0.341-x) M (0.220-x) M (2x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[HBr]^2}{[Br_2]\times [H_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHBr%5D%5E2%7D%7B%5BBr_2%5D%5Ctimes%20%5BH_2%5D%7D)
we are given : (0.341-x) = 0.268 M
x= 0.073 M
Thus equilibrium concentration of = (0.220-x) M = (0.220-0.073) M = 0.147 M
![[Br_2]=\frac{moles}{volume}\\0.147=\frac{xmole}{2.00L}\\\\x=0.294 mole](https://tex.z-dn.net/?f=%5BBr_2%5D%3D%5Cfrac%7Bmoles%7D%7Bvolume%7D%5C%5C0.147%3D%5Cfrac%7Bxmole%7D%7B2.00L%7D%5C%5C%5C%5Cx%3D0.294%20mole)
Thus there are 0.294 mol of present in the reaction vessel.
2 Na + SO3 = Na2SO3
Na = 1
SO3 = 1
Na2SO3 = 1
Na = 2
SO3 = 1
Na2SO3 = 1
I hope this helps!!
The increase of the boling point of a solution is a colligative property.
The formula for the increase of the normal boiling point of water is:
ΔTb = Kb * m
Where m is the molallity of the solution and Kb is the molal boiling constant in °C/mol.
ΔTb = 0.51 °C / m * 0.100 m = 0.051 °C.
So, the new boiling temperature is Tb = 100°C + 0.051°C = 100.051 °C.
Answer: 100.051 °C
Answer:
Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals.
