Answer: I think probaly wrong so probaly dont
listen to me
Explanation:
The compound that would be most reactive is Ethyne (answer A)
<u><em> explanation</em></u>
- Ethyne is the most unsaturated among the four compounds ( <em> it has a triple bond between the two carbon atoms) .</em>
- The triple bond in ethyne is made up of 1 sigma bond and 2π bond.
- <em>The 2π bond are weaker and can easily break which make Ethyne more reactive than Ethene, methane and Ethane.</em>
Answer:
Consequently, what happens when gas obtained by heating slaked lime and ammonium chloride is passed through copper sulphate solution? The HCl in the gas mixture will form hydrochloric and the H+ will react with some of the NH3(aq), forming NH4^+, and with some of the SO4^2-, forming HSO4^-.
Answer:
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Explanation:
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq).
Silver chromate is the salt of a strong base (AgOH) and a weak acid (H₂CrO₄).
HCrO₄⁻ is an even weaker acid than H₂CrO₄, so CrO₄²⁻ is a strong base.
Any added H⁺ will immediately combine with the chromate ions according to the reaction
H⁺ + CrO₄²⁻ ⟶ HCrO₄⁻
thereby removing chromate ions from solution.
According to Le Châtelier's Principle, more silver chromate will dissolve to replace the chromate ions that the H⁺ removes.
The overall equation for the reaction is
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + <em>CrO₄²⁻(aq)
</em>
<u>H⁺(aq) + </u><em><u>CrO₄²⁻(aq)</u></em><u> ⟶ HCrO₄⁻(aq)
</u>
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Answer:
Explanation:
a).
conc of Ca²⁺ =0.0025 M
pCa = -log(0.0025) = 2.6
logK,= 10.65 So lc = 4.47 x 10.
Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is 
=0.81
So the Conditional Formation constant= 
=0.81x 4.47 x10¹⁰
=3.62x10¹⁰
b)
At Equivalence point:
Ca²⁺ forms 1:1 complex with EDTA At equivalence point,
Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol
Number of moles of EDTA= 0.125 mol
Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL
V e= 25.00 mL
At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.
![[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M](https://tex.z-dn.net/?f=%5BCaY%5E%7B2-%7D%5D%20%3D%20%5Cfrac%7BInitial%2Cmoles%2Cof%2C%20Ca%5E%7B2%2B%7D%7D%7BTotal%2CVolume%7D%20%3D%20%5Cfrac%7B0.125mol%7D%7B%2850.00%2B25.00%29mL%7D%20%3D%200.001667M)
Ca²⁺ + Y⁴ ⇄ CaY²⁻
Initial 0 0 0.001667
change +x +x -x
equilibrium x x 0.001667 - x
![{K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\](https://tex.z-dn.net/?f=%7BK%5E%27%7D_f%20%3D%20%5Cfrac%7B%5BCaY%5E%7B2-%7D%5D%7D%7B%5BCa%5E%7B2%2B%7D%5D%5BY%5E4%5D%7D%3D%5Cfrac%7B0.001667-x%7D%7Bx.x%7D%20%3D%5Cfrac%7B0.001667-x%7D%7Bx%5E2%7D%5C%5C%5C%5Cx%5E2%20%3D%20%5Cfrac%7B0.001667-x%7D%7B%7BK%5E%27%7D_f%7D%5C%5C%20%5C%5C)
x = 2.15×10⁻⁷
[Ca+2] = 2.15x10⁻⁷ M
pca = —log(2 15x101= 6.7
