Answer:
![pH=-1.37](https://tex.z-dn.net/?f=%20pH%3D-1.37)
Explanation:
We are given that 25 mL of 0.10 M
is titrated with 0.10 M NaOH(aq).
We have to find the pH of solution
Volume of ![CH_3COOH=25mL=0.025 L](https://tex.z-dn.net/?f=CH_3COOH%3D25mL%3D0.025%20L)
Volume of NaoH=0.01 L
Volume of solution =25 +10=35 mL=![\frac{35}{1000}=0.035 L](https://tex.z-dn.net/?f=%5Cfrac%7B35%7D%7B1000%7D%3D0.035%20L)
Because 1 L=1000 mL
Molarity of NaOH=Concentration OH-=0.10M
Concentration of H+= Molarity of
=0.10 M
Number of moles of H+=Molarity multiply by volume of given acid
Number of moles of H+=
=0.0025 moles
Number of moles of
=0.001mole
Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles
Concentration of H+=![\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L](https://tex.z-dn.net/?f=%5Cfrac%7B0.0015%7D%7B0.035%7D%3D4.28%5Ctimes%2010%5E%7B-2%7D%20m%2FL)
pH=-log [H+]=-log [4.28
]=-log4.28+2 log 10=-0.631+2
![pH=-1.37](https://tex.z-dn.net/?f=%20pH%3D-1.37)
Answer:
I believe the answer is A: *It is the simplest form of matter" not 100% sure but I think that's correct
Explanation:
Answer:
3.81 g Pb
Explanation:
When a lead acid car battery is recharged, the following half-reactions take place:
Cathode: PbSO₄(s) + H⁺ (aq) + 2e⁻ → Pb(s) + HSO₄⁻(aq)
Anode: PbSO₄(s) + 2 H₂O(l) → PbO₂(s) + HSO₄⁻(aq) + 3H⁺ (aq) + 2e⁻
We can establish the following relations:
- 1 A = 1 c/s
- 1 mole of Pb(s) is deposited when 2 moles of e⁻ circulate.
- The molar mass of Pb is 207.2 g/mol
- 1 mol of e⁻ has a charge of 96468 c (Faraday's constant)
Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. The mass of lead deposited is:
![37.0s.\frac{96.0c}{s} .\frac{1mole^{-} }{96468c} .\frac{1molPb}{2mole^{-} } .\frac{207.2gPb}{1molPb} =3.81gPb](https://tex.z-dn.net/?f=37.0s.%5Cfrac%7B96.0c%7D%7Bs%7D%20.%5Cfrac%7B1mole%5E%7B-%7D%20%7D%7B96468c%7D%20.%5Cfrac%7B1molPb%7D%7B2mole%5E%7B-%7D%20%7D%20.%5Cfrac%7B207.2gPb%7D%7B1molPb%7D%20%3D3.81gPb)
The nonmetal elements have a negative charge.