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Deffense [45]
3 years ago
10

Lakes that have been acidified by acid rain can be neutralized by limiting the addition of limestone how much limestone is requi

red to completely neutralize a 4.3 billion liter lake with a ph of 5.5
Chemistry
1 answer:
svet-max [94.6K]3 years ago
4 0
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
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3 years ago
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The rate constant of a reaction is 4.7×10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C?
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Answer:

k is 3,18*10⁻² s⁻¹ at 75°C

Explanation:

following Arrhenius equation:

k= k₀*e^(-Ea/RT)

where k= rate constant , k₀= frequency factor , Ea= activation energy , R= universal gas constant T=absolute temperature

then for T₁=25°C =298 K

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k₂= k₀*e^(-Ea/RT₂)

dividing both equations

k₂/k₁= e^(-Ea/RT₂+Ea/RT₁ )

k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )]

replacing values

k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )] = 4,7*10⁻³ s⁻¹ *e^[-33.6*1000 J/mol /8.314 J/molK*(1/ 348 K -1/298 K )] = 3,18*10⁻² s⁻¹

thus k is 3,18*10⁻² s⁻¹ at 75°C

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