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
Answer:
see below
Explanation:
for A + 2B => Products ...
Rate Law => Rate =k[A][B]ˣ
As shown in expression, A & B are included, C is not.
Answer:
okkkkk
Explanation:
the balanced equation is
2NaBr +Cl2 =2NaCl + Br2
can you mark me as a brainliest
Answer:
A is a physical change, no atoms are being lost its still in it original form
B is a chemical change, it is losing atoms and changing into a new substance
Explanation:
Remark
The question with these kind of problems is "Which R do you use?" That's where dimensional analysis is so handy. You must look at the units of the givens and choose your R accordingly. You'll see how that works in a moment.
You need to list the givens along with their units and in this case the property you want to solve for. You need all that to determine the R value
Givens
n = 0.25 moles
T = 35°C = 35 + 273.15 = 308.15°K
V = 6.23 L
Pressure = P in kPa
Which R
The units of the R you want has to have units of moles, kPa, °K and liters
The R that you want is 8.314
<em><u>Formula</u></em>
PV = nRT
P 6.23 = 0.25 * 8.314 * 308.15 Combine the left
P*6.23 = 640.5
P = 640.5/6.23 = 102.81 The answer should be 100 kpA of 1.0 * 10^2 kPa
because the number of moles has only 2 sig digs.
But if sig digs are not a problem 102.8 is likely close enough.
Second Question
You are going to have to clean up the numbers. I think I've got only 1 chance at this. The partial pressures of the 2 gases will add up to the total pressure. So the total pressure was 100 approx and the water vapor was 3.36 kPa. The difference is
Total = air + water vapor
100.18 = air + 3.36 about Subtract 3.36 from both sides.
100.18 - 3.36 = 96.82 about. Pick the answer that is closest to that. I'll clean up the numbers if I can.
Answer C
