The chemical reaction would be:
<span>CuSO4(aq) + NaOH(aq) = Cu(OH)2(s) + Na2SO4(aq)
One observation would be the formation of a precipitate since one of the products is not readily soluble to aqueous solution. A formation of a blue precipitate will be observed.</span>
Answer: Thus the value of 
is 110.25
Explanation:
Initial moles of 
= 0.500 mole
Initial moles of 
= 0.500 mole
Volume of container = 1 L
Initial concentration of 
Initial concentration of 
equilibrium concentration of 
[/tex]
The given balanced equilibrium reaction is,
Initial conc. 0.500 M 0.500 M 0 M
At eqm. conc. (0.500-x) M (0.500-x) M (2x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[IBr]^2}{[Br_2]\times [I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BBr_2%5D%5Ctimes%20%5BI_2%5D%7D)
we are given : 2x = 0.84 M
x= 0.42
Now put all the given values in this expression, we get :
Thus the value of the equilibrium constant is 110.25
Answer:
14 electrons
Explanation:
The total number of electrons that can occupy the f sublevel is 14.
The d sublevel can have 10 electrons.
The p sublevel can have 6 electrons
The s sublevel can have 2 electrons.
We can find the mass of ammonia using the ideal gas law equation,
PV = nRT
where
P - pressure - 2.15 atm x 101 325 = 2.18 x 10⁵
V - volume - 3.00 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 15.0 °C + 273 = 288 K
substituting these values in the equation
2.18 x 10⁵ Pa x 3.00 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
n = 0.273 mol
number of moles of NH₃ is 0.273 mol
molar mass of NH₃ - 17.0 g/mol
mass pf ammonia present - 0.273 mol x 17.0 g/mol = 4.64 g
mass of NH₃ present is 4.64 g
