From the given pH, we calculate the concentration of H+:
[H+] = 10^-pH = 10^-5.5
We then use the volume to solve for the number of moles of H+:
moles H+ = 10^-5.5M * 4.3x10^9 L = 13598 moles
From the balanced equation of the neutralization of hydrogen ion by limestone written as
CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2CO3(aq)
we use the mole ratio of limestone CaCO3 and H+ from their coefficients, which is 1 mole of CaCO3 is to react with 2 moles of H+, to compute for the mass of the limestone:
mass CaCO3 = 13598mol H+(1mol CaCO3/2mol H+)
(100.0869g CaCO3/1mol CaCO3)(1kg/1000g)
= 680 kg
Answer:
Wave particle duality is most apparent in analyzing the motion of. an electron. A photon of which electromagnetic radiation has the most energy.
The answer is B.
occur gradually over millions of years.
We balance the given reactions above by following the rules in balancing redox reactions in acidic or basic solutions. Balance the atoms aside from the O and H atoms. Then we balance the Os and Hs by adding H2O or H+. Finally, we balance the total charge of the reactant and product by adding e-. We do as follows:
<span>A) H2O2 + Fe 2+ ---> Fe 3+ + H2O (in the acidic solution)
</span><span> 2H+ + </span>H2O2 + Fe 2+ ---> Fe 3+ + 2H2O
e- + 2H+ + H2O2 + Fe 2+ ---> Fe 3+ + 2H2O
<span>
C) CN- + MnO4- ---> CNO- +MnO2 (in basic solution)
</span> CN- + MnO4- ---> CNO- +MnO2 + H2O
2H+ + CN- + MnO4- ---> CNO- +MnO2 + H2O
2OH- + 2H+ + CN- + MnO4- ---> CNO- +MnO2 + H2O + 2OH-
2H2O + CN- + MnO4- ---> CNO- +MnO2 + H2O + 2OH-
e- + H2O + CN- + MnO4- ---> CNO- +MnO2 + 2OH-
<span>
E) S2O2/3- + I2 ---> I- + S4O2/6- (in acidic solution)
2</span>S2O2/3- + I2 ---> 2I- + S4O2/6-
4H+ + 2S2O2/3- + I2 ---> 2I- + S4O2/6- + 2H2O
6e- + 4H+ + 2S2O2/3- + I2 ---> 2I- + S4O2/6- + 2H2O
This is an oxidation reaction. The balanced equation is as follows:
6H₂O + N₂ → 2NO₃⁻+ 12H⁺ + 6e⁻
Rules to balance redox reaction in acidic medium
- Write the given equation in ionic form
- Identify elements undergoing oxidation ( charge increase, O.N inc) and reduction (charge dec, O.N dec)
- Break the equation into two halfs
- Balance the half equations
A. Balance all other atoms except Oxygen and hydrogen
B. Balance oxygen by adding H2O to the side deficient in oxygen
C. Balance hydrogen by adding H+ ions
D. Balance charge by adding electrons
5. Add the two half such that electrons gets cancelled
Oxidation number of N in N2 is 0 while in NO₃⁻, it is +5. Thus there is an increase in oxidation number, thus oxidation is taking place.
N₂(g) → NO₃⁻(aq)
N₂ → 2NO₃⁻
6H₂O + N₂ → 2NO₃⁻
6H₂O + N₂ → 2NO₃⁻+ 12H⁺
- balance charge. 0 charge on left, -6 and + 12 on right. add 6e⁻ on right to balance.
6H₂O + N₂ → 2NO₃⁻+ 12H⁺ + 6e⁻
Thus we can conclude that since there is increase in oxidation number, oxidation is taking place.
learn more about balancing redox reactions at brainly.com/question/10203480
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