Answer:
A. Acidic
B. Neutral
C. Basic
D. Acidic
E. Basic
Explanation:
The reaction of salts with water to give aqueous solutions which are either acidic, basic or neutral is known as salt hydrolysis. Salts that are produced from the reaction of strong acids and weak bases give acidic solutions. Salts that are produced the reaction between weak acids and strong bases give basic solutions. Neutral solutions are obtained from salts produced bynthe reaction of strong acids and strong bases.
The hydrolysis of the given salts are shown below:
A. AlCl₃ + 3 H₂O ---> 3 HCl + Al(OH)₃
The aqueous solution produced will be acidic since HCl is a strong acid while Al(OH)₃ is a weak base.
B. NaBr + H₂O ----> HBr + NaOH
The aqueous solution produced will be neutral since HBr is a strong acid and NaOH is strong base as well.
C. NaClO + H₂O ----> HClO + NaOH
The aqueous solution produced will be basic since HClO is a weak acid while NaOH is a strong base.
D. CH₃NH₃NO₃ + H₂O ----> HNO₃ + CH₃NH₂ + H₂O
The aqueous solution produced will be acidic since HNO₃ is a strong acid while CH₃NH₂ is a weak base.
E. Na₂SO₃ + 2 H₂O ----> H₂SO₃ + 2 NaOH
The aqueous solution produced will be basic since H₂SO₃ is a weak acid while NaOH is a strong base.
It’s [Xe]7s^2 5f^14 (first one) hopefully i could help!!
Answer: green house affect is the sun on plants it is transfering energy to the plants to help them grow same with every plant, tree, fruit tree. everything needs sun.
Explanation: look up there.
Answer : The value of 
for this reaction is, 
Explanation :
The given chemical reaction is:
Now we have to calculate value of 
.
![\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5Bn_%7BHCH_3CO_2%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28HCH_3CO_2%28g%29%29%7D%5D-%5Bn_%7BCH_3OH%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CH_3OH%28g%29%29%7D%2Bn_%7BCO%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CO%28g%29%29%7D%5D)
where,
= Gibbs free energy of reaction = ?
n = number of moles
= -389.8 kJ/mol
= -161.96 kJ/mol
= -137.2 kJ/mol
Now put all the given values in this expression, we get:
![\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5B1mole%5Ctimes%20%28-389.8kJ%2Fmol%29%5D-%5B1mole%5Ctimes%20%28-163.2kJ%2Fmol%29%2B1mole%5Ctimes%20%28-137.2kJ%2Fmol%29%5D)
The relation between the equilibrium constant and standard Gibbs, free energy is:
where,
= standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol
R = gas constant = 8.314 J/L.atm
T = temperature = 
= equilibrium constant = ?
Now put all the given values in this expression, we get:
Thus, the value of for this reaction is,
