Answer:
HNO₃ + NaOH ---> NaNO₃ + H₂O
Explanation:
This reaction appears to be a double-displacement reaction. In these reaction, the cation of one compound is swapped with the cation of another.
As such, the hydrogen cation (H⁺) from HNO₃ is swapped with the sodium cation (Na⁺) of NaOH.
Luckily, all of the cations have a +1 charge and the anions have a -1 charge. This means that no coefficients are necessary to balance the reaction.
The <u>complete balanced </u>equation is:
HNO₃ + NaOH ---> NaNO₃ + H₂O
Answer is Mr and S as MgS ..
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
Answer:
ΔS = -661.0J/mol is the entropy change for the system
ΔS = -842J/mol.K is the entropy change for the surroundings
Explanation:
From the relationship between ΔG, T, ΔH and ΔS,
Mathematically, ΔG = ΔH - TΔS
TΔS = ΔH - ΔS
ΔS = ΔH - ΔS / T
but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)
plugging into the equation,
ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298
ΔS = -0.6610KJ/mol or in J.mol
ΔS = -661.0J/mol is the entropy change for the system
- For entropy change for the surroundings = ΔS = ΔH/T
- ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings
