Explanation:
See this l know natural only
An H atom is made up of a nucleus with a +1 charge, as well as a single electron. Therefore, the only positively charged ion possible has charge +1.
I hope this helps!
(a) Iron (iii) sulphate:
From the periodic table:
mass of iron = 55.845 grams
mass of sulphur = 32.065 grams
mass of oxygen = 16 grams
Iron (iii) sulphate has the formula: Fe2(SO4)3
molar mass = 2(55.845) + 3(32.065) + 3(4)(16) = 399.885 grams
(b) Sodium hydroxide:
From the periodic table:
mass of sodium = 22.989 grams
mass of oxygen = 16 grams
mass of hydrogen = 1 gram
Sodium hydroxide has the formula: NaOH
molar mass = 22.989 + 16 + 1 = 39.989 grams
(c) Barium carbonate
From the periodic table:
mass of barium = 137.327 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Barium carbonate has the formula: BaCO3
molar mass = 137.327 + 12 + 3(16) = 197.327 grams
(d) ammonium nitrate:
From the periodic table:
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
Ammonium nitrate has the formula: NH4NO3
molar mass = 14 + 4(1) + 14 + 3(16) = 80 grams
(e) Lead (iv) oxide
From the periodic table:
mass of lead = 207.2 grams
mass of oxygen = 16 grams
Lead (iv) oxide has the formula: PbO2
molar mass = 207.2 + 2(16) = 239.2 grams
From the above calculations, we can see that:
Iron (iii) sulphate has the greatest mass.
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
