<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.
<u>Explanation:</u>
The chemical formula of carbonic acid is 
. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water
The chemical equation for the first dissociation of carbonic acid follows:
The expression of first equilibrium constant equation follows:
![Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}](https://tex.z-dn.net/?f=Ka_1%3D%5Cfrac%7B%5BH%5E%2B%5D%5BHCO_3%5E%7B-%7D%5D%7D%7B%5BH_2CO_3%5D%7D)
The chemical equation for the second dissociation of carbonic acid follows:
The expression of second equilibrium constant equation follows:
![Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}](https://tex.z-dn.net/?f=Ka_2%3D%5Cfrac%7B%5BH%5E%2B%5D%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BHCO_3%5E-%5D%7D)
Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.
Answer:
ΔU=-369.2 kJ/mol.
Explanation:
We start from the equation:
Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.
If Δ(PV) were calculated by ideal gas law,
PV=nRT
Δ(PV)=RTΔn.
Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.
So, for this case, ΔH=ΔU.
The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:
ΔU=-369.2 kJ/mol.
Answer:
The molecular formula of cacodyl is C₄H₁₂As₂.
Explanation:
<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:
- 209.96 g * 22.88/100 = 48.04 g C
- 209.96 g * 5.76/100 = 12.09 g H
- 209.96 g * 71.36/100 = 149.83 g As
Now we convert those masses into moles:
- 48.04 g C ÷ 12 g/mol = 4.00 mol C
- 12.09 g H ÷ 1 g/mol = 12.09 mol H
- 149.83 g As ÷ 74.92 g/mol = 2.00 mol As
Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.
Answer:
by using ideal gas law
Explanation:
ideal gas law:
PV=nRT
where:
P is pressure measured in Pascal (pa)
V is volume measured in letters (L)
n is number of moles
R is ideal gas constant
T is temperature measured in Kelvin (K)
by applying the given:
P(initial) V(initial)=nRT(initial)
P(final) V(final)=nRT(final)
nR is constant in both equations since same gas
then,
P(initial) V(initial) / T(initial) = P(final) V(final) / T(final)
then by crossing multiply both equations
V (final)= { (P(initial) V(initial) / T(initial)) T(final) } /P (final)
P(initial)=P(final)= 1 atm = 101325 pa
V(initial)= 6 L
T(initial) = 28°c = 28+273 kelvin
T(final) = 39°c = 39+273 kelvin
by substitution
V(final) = 6.21926 L
Answer:
Option A
Explanation:
Number of millimoles of Na3PO4 = 1 × 100 = 100
Number of millimoles of AgNO3 = 1 × 100 = 100
When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion
When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-
As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible
Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-
And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-
∴ Increasing order of concentration will be PO43- < NO3- < Na+
