The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M
<h3>Balanced equation </h3>
H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 1
- The mole ratio of the base, KOH (nB) = 3
<h3>How to determine the molarity of H₃PO₄ </h3>
- Volume of acid, H₃PO₄ (Va) = 10.2 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
- Volume of base, Ca(OH)₂ (Vb) = 53.5 mL
- Molarity of acid, H₃PO₄ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 10.2) / (0.2 × 53.5) = 1 / 3
(Ma × 10.2) / 10.7 = 1 / 3
Cross multiply
Ma × 10.2 × 3 = 10.7
Ma × 30.6 = 10.7
Divide both side by 30.6
Ma = 10.7 / 30.6
Ma = 0.35 M
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Answer:
c
Explanation:
because potential energy is created with the position of a body when you pick up something the thing will have stored energy which is transferred from you so the stored energy is potential energy created by the position of a body that when you see over your head you scared of that thing not to bit you because it has stored energy if you see the thing already failed on the ground you wouldn't scared of that because the enery is already gone
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Where are the answers that apply .. hopes this help :)
Answer:
The reaction will be non spontaneous at these concentrations.
Explanation:
Expression for an equilibrium constant 
:
![K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BAg%5E%2B%5D%5BBr%5E-%5D%7D%7B%5BAgCl%5D%7D%3D%5Cfrac%7B%5BAg%5E%2B%5D%5BBr%5E-%5D%7D%7B1%7D%3D%5BAg%5E%2B%5D%5BBr%5E-%5D)
Solubility product of the reaction:
![K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BBr%5E-%5D%3DK_c%3D7.7%5Ctimes%2010%5E%7B-13%7D%20)
Reaction between Gibb's free energy and equilibrium constant if given as:
![\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-2.303%5Ctimes%208.314%20J%2FK%20mol%5Ctimes%20298%20K%5Ctimes%20%5Clog%5B7.7%5Ctimes%2010%5E%7B-13%7D%5D)
Gibb's free energy when concentration ![[Ag^+] = 1.0\times 10^{-2} M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%20%3D%201.0%5Ctimes%2010%5E%7B-2%7D%20M)
and ![[Br^-] = 1.0\times 10^{-3} M](https://tex.z-dn.net/?f=%5BBr%5E-%5D%20%3D%201.0%5Ctimes%2010%5E%7B-3%7D%20M)
Reaction quotient of an equilibrium = Q
![Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}](https://tex.z-dn.net/?f=Q%3D%5BAg%5E%2B%5D%5BBr%5E-%5D%3D1.0%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%201.0%5Ctimes%2010%5E%7B-3%7D%20M%3D1.0%5Ctimes%2010%5E%7B-5%7D)
![\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])](https://tex.z-dn.net/?f=%5CDelta%20G%3D69.117%20kJ%2Fmol%2B%282.303%5Ctimes%208.314%20Joule%2Fmol%20K%5Ctimes%20298%20K%5Ctimes%20%5Clog%5B1.0%5Ctimes%2010%5E%7B-5%7D%5D%29)
- For reaction to spontaneous reaction:
. - For reaction to non spontaneous reaction:
.
Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations
