That acid rain has lowered the ph of a particular lake to ph 5.0. the hydroxide ion concentration of this lake is
1 × 10^-9 mol of hydroxide ions per liter of lake water
pH + pOH = 14
pOH = 14 - 5
pOH = 9.00
pOH = -log [OH⁻]
[OH⁻] = 10^-pOH
[OH⁻] = 
mol
Therefore,
the hydroxide ion concentration of this lake is 1 × 10^-9 mol
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(NH4)2CO3(aq)+ZnCl2 (aq)-----> ZnCO3(s)+2NH4Cl(aq)
Zn²⁺ +CO3²⁻-----> ZnCO3
Given, mass of isotope 1 = 39.06 amu
% abundance of isotope 1 = 10.00%
mass of isotope 2 = 42.14 amu
% abundance of isotope 2 = 90.00%
Average atomic mass
= [ (mass of isotope 1 x % abundance of isotope 1 ) + (mass of isotope 2 x % abundance of isotope 2 ) ]/ 100
= [(39.06 x 10) + (42.14 x 90)]/100
= (390.6 + 3792.6) / 100
= 4183.2/100
= 41.8 (to the nearest tenth)
Therefore, the weighted atomic mass of element Q is 41.8 amu.
From the calculation, the partial pressure of Cl2 is 1.98 atm. Option B
<h3>What is the equilibrium constant?</h3>
The term equilibrium constant refers to the value that shows us the extent to which reactants are converted into products. We know that the equilibrium constant can only be obtained by the use of the equilibrium partial pressures of each of the species.
Thus we have;
K = p PCl3 . pCl2/pPCl5
p PCl3 = partial pressure of PCl3
pCl2 = partial pressure of Cl2
pPCl5 = partial pressure of PCl5
K = equilibrium constant
Substituting values;
1.05 = 0.463 * pCl2/0.875
pCl2 = 1.05 * 0.875/0.463
pCl2 = 1.98 atm
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