Answer:
See explanation
Explanation:
The oxides or hydrides are formed by exchange of valency between the two atoms involved. The group of the atom bonded to oxygen or hydrogen in the binary compound can be deduced by considering the subscript attached to the oxygen or hydrogen atom.
Now let us take the journey;
R2O3- refers to an oxide of a group 13 element, eg Al2O3
R2O - refers to an oxide of group a group 1 element e.gNa2O
RO2 - refers to an oxide of a group 14, 15 or 16 element such as CO2, NO2 or SO2
RH2 - refers to the hydride of a group 12 element Eg CaH2
R2O7 - refers to an oxide of a group 17 element E.g Cl2O7
RH3- refers to a hydride of a group 13 element E.g AlH3
Answer:
pH = 9,32
Explanation:
The compound with 2 ionizable groups has the following equilibriums:
H₂M ⇄ HM⁻ + H⁺ pka = 6,2
HM⁻ ⇄ M²⁻ + H⁺ pka = 9,5
The reaction of M²⁻ with HCl is:
M²⁻ + HCl → HM⁻ + Cl⁻
The moles of M²⁻ are:
0,100L×1,0M = 0,1moles
And moles of HCl are:
0,060L×1,0M = 0,06moles
That means that moles of M²⁻ will be 0,1-0,06 = 0,04mol and moles of HM⁻ will be the same than HCl, 0,06mol
Using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [M²⁻] / [HM⁻]
Replacing:
pH = 9,5 + log₁₀ [0,04] / [0,06]
<em>pH = 9,32</em>
<em></em>
I hope it helps!
28. unbalanced because in reactant side the Aluminium is 1 and oxygen is 2 but in the product side the Aluminium is 2 and oxygen is 3 it is not equal so it is unbalanced equation
29. unbalanced because the oxygen atom are not equal in the reactant and the product side
Follow me pls!!
Answer: The correct option is, (C) 0.53
Explanation:
The given chemical reaction is:
The rate of the reaction for disappearance of A and formation of C is given as:
![\text{Rate of disappearance of }A=-\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DA%3D-%5Cfrac%7B1%7D%7B9%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D)
Or,
![\text{Rate of formation of }C=+\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DC%3D%2B%5Cfrac%7B1%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
where,
= change in concentration of C = 1.33 M
= change in time = 4.5 min
Putting values in above equation, we get:
![\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}=\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
![\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B9%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
![\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{1.33M}{4.5min}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B9%7D%7B5%7D%5Ctimes%20%5Cfrac%7B1.33M%7D%7B4.5min%7D)
![\frac{\Delta [A]}{\Delta t}=0.53M/min](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D0.53M%2Fmin)
Thus, the decrease in A during this time interval is, 0.53
Answer:
Density of unit cell ( rhodium) = 12.279 g/cm³
Explanation:
Given that:
The radius (r) of a rhodium atom = 135 pm
The atomic mass of rhodium = 102.90 amu
For a face-centered cubic unit cell,
where;
a = edge length.
Making "a" the subject of the formula:
a = 381.8 pm
to cm, we get:
a = 381.8 × 10⁻¹⁰ cm
However, recall that:
where;
mass of unit cell = mass of atom × numbers of atoms per unit cell
Also;
Recall also that number of atoms in a unit cell for a face-centered cubic = 4
So;
mass of unit cell = 6.83380375 × 10⁻²² g
Density of unit cell ( rhodium) = 12.279 g/cm³
