Balance the chemical equation given below, and determine the number of grams of MgO that are needed to produce 20.0 g of Fe2O3.
1 answer:
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The half life for C14 is 5730 years.
We assume that Carbon 14/ Carbon 12 ratio was steady for living organisms over time, the problem is actually telling us that
= 0.0725 = 
ˣ
Take the natural logarithm and In on both sides.
ln(0.725) = lnˣ
= - 0.3216 = xln ( = -0.6931x.
So x = (-.3216) / (-0.6931) = 0.464
or
t/t₁/₂ = 0.464
So t = 0.464 x t₁/₂ = 0.464 * 5730 yrs = 2660 years.
We assume that Carbon 14/ Carbon 12 ratio was steady for living organisms over time, the problem is actually telling us that
Take the natural logarithm and In on both sides.
ln(0.725) = ln
= - 0.3216 = xln (
So x = (-.3216) / (-0.6931) = 0.464
or
t/t₁/₂ = 0.464
So t = 0.464 x t₁/₂ = 0.464 * 5730 yrs = 2660 years.
Answer:
Check the explanation
Explanation:
When,
pH = -log[H+] = 3.30
[H+] =
=
=
When,
pH = -log[H+] = 10.15
[H+] =
Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = ; Ka6 =
=
=