Answer:
There are 1.51 x 1024 molecules of carbon dioxide in 2.50 moles of carbon dioxide.
Explanation:
The atomic radius increases as you would go down a particular group on the periodic table of elements. This is because along with a greater number of protons, there would also be electrons as well, and thus the need of electron shells surrounding the atom would also be required, to compensate for the more electrons, as according to the bohr model, each shell contains 8 electrons in its electron shell. Thus the distance from the nucleus to the outermost shell increases, the atomic radius.
Answer:
k = [F2]² [PO]² / [P2] [F2O]²
Explanation:
In a chemical equilibrium, the equilibrium constant expression is written as the ratio between the molar concentration of the products over the molar concentration of the reactants. Each species powered to its reaction coefficient. For the equilibrium:
P2(g) + 2F2O(g) ⇄ 2PO(g) + 2F2(g)
The equilibrium constant, k, is:
k = [F2]² [PO]² / [P2] [F2O]²
Answer:
see explanation below
Explanation:
First to all, this is a redox reaction, and the reaction taking place is the following:
2KMnO4 + 3H2SO4 + 5H2O2 -----> 2MnSO4 + K2SO4 + 8H2O + 5O2
According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:
MhVh = MpVp
h would be the hydrogen peroxide, and p the permanganate.
But like it was stated before, the mole ratio is 5:2 so:
5MhVh = 2MpVp
Replacing moles:
5nh = 2MpVp
Now, we just have to replace the given data:
nh = 2MpVp/5
nh = 2 * 1.68 * 0.0213 / 5
nh = 0.0143 moles
Now to get the mass, we just need the molecular mass of the peroxide:
MM = 2*1 + 2*16 = 34 g/mol
Finally the mass:
m = 0.0143 * 34
m = 0.4862 g
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M