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.
Answer:
mass of HCl = 243.5426 grams
Explanation:
1- we will get the mass of the reacting gold:
volume of gold = length * width * height
volume of gold = 3.2 * 3.8 * 2.8 = 34.048 cm^3 = 34.048 ml<span>
density = mass / volume
Therefore:
mass = density * volume
mass of gold = </span>19.3 * 34.048 = 657.1264 grams
2- we will get the number of moles of the reacting gold:
number of moles = mass / molar mass
number of moles = 657.1264 / 196.96657
number of moles = 3.3362 moles
3- we will get the number of moles of the HCl:
First, we will balanced the given equation. The balanced equation will be as follows:
Au + 2HCl ......> AuCl2 + H2
This means that one mole of Au reacts with 2 moles of HCl.
Therefore 3.3362 moles will react with 2*3.3362 = 6.6724 moles of HCL
4- we will get the mass of the HCl:
From the periodic table:
molar mass of H = 1 gram
molar mass of Cl = 35.5 grams
Therefore:
molar mass of HCl = 1 + 35.5 = 36.5 grams/mole
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass of HCl = 6.6724 * 36.5
mass of HCl = 243.5426 grams
Hope this helps :)
If 20.0 grams of ice takes about 6680j...then 40.0 grams of ice would take ___ .
