The total number of lone pairs of electrons in N2O3 is 8.
Answer:
1. (NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)
2. Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)
Explanation:
The dissociation of ammonium sulphide, (NH₄)₂S when dissolved in water is given in the equation below:
(NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)
However very little S²- ions are present in solution due to the very basic nature of the S²- ion (Kb = 1 x 105).
The ammonium ion being a better proton donor than water, donates a proton to sulphide ion to form hydrosulphide ion which exists in equilibrium with aqueous ammonia.
S²- (aq) + NH₄+ (aq) ⇌ SH- (aq) + NH₃ (aq)
Aqueous solutions of ammonium sulfide are smelly due to the release of hydrogen sulfide and ammonia, hence, their use in making stink bombs.
2. The reaction between aluminium nitrate and sodium phosphatein aqueous solution is a double decomposition reaction whish results in the precipitation of insoluble aluminium phosphate. The equation of the reaction is given below :
Al(NO₃)₃ (aq) + Na₃PO₄ (aq) ----> AlPO₄ (s) + 3 NaNO₃ (aq)
The net ionic equation is given below:
Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)
Carbon dioxide dissolves faster
Answer:
In conclussion, 0.60 moles of HCOOH contains the greatest mass of O
Explanation:
Let's make some rules of three, to solve this problem:
1 mol of ethanol has 2 moles of C, 6 moles of H, and 1 mol of oxygen
Therefore, 0.75 moles of ethanol must have 0.75 mol of oyxgen
Let's convert the moles to mass → 0.75 mol . 16 g/ 1 mol = 12 g
1 mol of formic acid has 2 moles of H, 1 mol of C and 2 mol of oxygen
0.60 moles of formic acid must have (0.6 .2) / 1 = 1.2 mol of O
If we convert the amount to mass → 1.2 mol . 16 g/ 1mol = 19.2 g
1 mol of water has 1 mol of oyxgen
Therefore, we have 1 mol of oxygen with a mass of 16 g.
In conclussion, 0.60 moles of HCOOH contains the greatest mass of O
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>
