Density is defined as mass per unit volume so even when you cut an object in half unit volume does not change so each part would have a different density even if it’s cut into the same pieces
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:
1.09 grams
Explanation:
According to the following chemical equation:
HF + NaNO₃ -> HNO₃ + NaF
1 mol of hydrogen fluoride (HF) produces 1 mol of sodium fluoride (NaF). Thus, we first convert from mol to grams by using the molar mass (MM) of each compound:
MM(HF)= (1 g/mol x 1 H) + (19 g/mol x 1 F) = 20 g/mol HF
1 mol HF x 19.9 g/mol HF = 20 g
MM(NaF) = (23 g/mol x 1 Na) + (19 g/mol x 1 F) = 42 g/mol NaF
1 mol NaF x 42 g/mol NaF = 42 g
Thus, from 20 g of HF are produced 42 g of NaF ⇒ 20 g HF/42 g NaF. We multiply this stoichiometric ratio by the mass of NaF produced to calculate the required mass of HF:
20 g HF/42 g NaF x 2.3 g NaF = 1.09 g HF
Therefore, 1.09 grams of HF are necessary to produce 2.3 g of NaF.
The percentage of glucose given is m/v. This means that the given percentage of volume consists of mass.
In this solution, percentage of glucose is 5.5% m/v.
This means that 5.5% of the volume is the mass of glucose.
Given volume is 285 mL.
Therefore mass of glucose is 5.50% of 285 mL = (5.5*285)/100
mass of glucose = 15.67 g
Hi!
Fuel-cell cars only release water and heat in the atmosphere.
Since they are only run by oxygen and hydrogen to produce electricity in the motor, they don't emit carbon dioxide or greenhouse gas as what other people think.
Basically, automotive engineers and scientist classified them as zero-emission.
