A. is decomposition so HCL = H2 + Cl2
not balanced cause hcl needs 2
2HCL = H2 + Cl2
balanced
b. Br2 + Al-i = AlBr3 + I2 single rep.
not balanced since br need 3 so watch carefully cause many changes needed
3Br2 + Al-i = AlBr3 + I2 not right is unbalanced so make it 2
3Br2 + Al-i = 2AlBr3 + I2 now left Al is unbal. so make 2 there
3Br2 + 2Ali = 2AlBr3 + I2
Balanced
C. Na + S = Na2S synthesis reaction is not bal. left Na needs 2
2Na + S = Na2S balanced.
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
Hey Brayden!
Rain showers had continued over a particular area for several days. Which of these weather factors is most likely responsible?
Answer - A Stationary front
Hope This Helps!
If you need anymore help let me know!
You have to find the stoichiometric ratio between AlCl₃ and BaCl₂. The common element between them is Cl. So, the ratio of Cl in BaCl₂ to AlCl₃ is 2/3. The molar mass of AlCl₃ is 133.34 g/mol. The solution is as follows:
Mass of AlCl₃ = (6 mol BaCl₂)(2 mol Cl/1 mol BaCl₂)(1 mol AlCl₃/3 mol Cl)(133.34 g/mol) = 533.36 g AlCl₃
