Look at the letters, 3 would be MMeters
<h3><u>Answer;</u></h3>
0.5 M HBr, pOH = 13.5 ; Has the lowest pH
<h3><u>Explanation;</u></h3>
From the question;
pH = -Log [OH]
or pH = 14 - pOH
Therefore;
For 0.5 M HBr
[H+] = 0.5 M
pH = - Log [0.5]
= 0.30
For; pOH = 13.5
pH = 14 - pOH
= 14 -13.5
= 0.5
For; 0.05 M HCl
pH = - log [H+]
[H+] = 0.05
pH = - Log [0.05]
= 1.30
For; pOH = 12.7
pH = 14 -pOH
= 14 -12.7
= 1.30
For; 0.005 M KOH,
pOH = - log [OH]
[OH-] = 0.005
pOH = - Log 0.005
= 2.30
pH = 14 - 2.30
= 11.7
For; pOH = 2.3
pH = 14 -pOH
= 14- 2.3
= 11.7
Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
According to periodic trends, bismuth has the greatest atomic radius.
They have hydrogens numbers wrong its 1.008 not 1.01
