We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375 0 0
Change -x +x +x
Equilibrium 0.0375-x x x
We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)
Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.
We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
Answer:
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Answer:
Enzyme Active Site and Substrate Specificity
There may be one or more substrates for each type of enzyme, depending on the particular chemical reaction. In some reactions, a single-reactant substrate is broken down into multiple products. In others, two substrates may come together to create one larger molecule.
They increase across each period, decrease down a group. As you go across a period the number of protons and increases. The positive nucleus then has a stronger attractive force on the electrons so it takes a larger amount of energy to remove an electron. As you go down a group the atoms are larger so the attractive force is weaker and it takes less energy to remove an electron.
A bond between two atoms that SHARE electrons
