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:
Explanation:
Use the gas equation: PV=nRT
P=pressure
V=Volume
R= gas constant of around 8.31 J/K/mol
T=temperature
n= number of moles
To find n, Rearrange:
n=PV/RT
102kPa= 102,000 kilo pascals which standard form is 102 x 10^3
Convert Celsius to kelvin, which you just add 273.15. So:
37+273.15=310.15 K round to a whole number is 310 K
Sub in all numbers to calculate the mol
n= 102 x 10^3 x 2.20 x 10^3/ 8.31 x 310 (cross out 10^3 as this will make a big number)
n=102 x 2.2/8.31 x 310 =0.087 mol
We know 1 g=29 moles
Multiply 29 moles with 0.087 to find the grams
29*0.087=2.523, which to one d.p is 2.5 g
Hence, the child's lung will hold 2.5 g of air.
Hope this helps you :)
Have a nice day!!
<em>mC: 12g/mol</em>
12g ------- 6,02·10²³ a.
6g --------- X
X = (6×<span>6,02·10²³)/12
</span><u>X = 3,01·10²³ atoms</u>
---->>> A :)
I think it would be solubility but I’m not sure
Answers:
<u>Option B. 4.</u>
Explanation:
The visible spectrum of light emitted by a sample of active or excited hydrogen atoms splits into four wavelengths that are basically four distinct levels of energy in the visible region of the spectrum. There four different wavelengths are 410 nm, 434 nm, 486 nm, and 656 nm. These show for colors according to the spectrum and wavelength violet, blue, green, and red, where 656 nm wavelength is most intense on the spectrum.
