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#t = "145.5 s"#
(For the record, I did this without looking at the answer. It's much more important you know HOW to do it.)
The first method is a very general approach as in radioactive decay problems (which are first-order processes!). The second method is from the kinetics unit in general chemistry.
METHOD 1
This can be done simply by knowing the definition of a half-life:
The time it takes for the concentration of a sample to halve.
Since we know that
#t_"1/2" = "43.80 s"# ,
consider the following recursive train of thought for first-order half-lives only.
and so on. Thus, we can write the current concentration as a function of the starting concentration:
#[A] = (1/2)^n [A]_0# where
#n# is the number of half-lives that passed.
The number of half-lives passed,
Thus,
#barul|stackrel(" ")(" "[A] = (1/2)^(t//t_"1/2") [A]_0" ")|#
Given that
#[A] = 0.1[A]_0# ,
and thus,
#(1/2)^(t//t_"1/2") = 0.1# .
To solve for
#ln(1/2)^(t//t_"1/2") = ln0.1#
#t/(t_"1/2")ln(1/2) = ln0.1#
Therefore, it will take this long for
#color(blue)(t) = t_"1/2"cdotln0.1/ln(1/2)#
#= "43.80 s" cdot (ln 0.1)/(ln (1/2))#
#=# #color(blue)("145.5 s")#
METHOD 2
The more usual method uses the half-life of a first-order reaction (refer to your textbook):
#t_"1/2" = (ln2)/k#
And thus, the rate constant is given by:
#k = (ln2)/t_"1/2"#
Via the first-order integrated rate law (refer to your textbook),
#ln[A] = -kt + ln[A]_0# ,
one can then solve for the time passed simply by knowing that
#ln(0.1[A]_0) - ln[A]_0 = -kt#
#ln((0.1cancel([A]_0))/(cancel([A]_0))) = -kt#
#t = ln0.1/(-k)#
#= (ln0.1)/(- (ln2)/(t_"1/2"))#
#= t_"1/2" cdot (ln0.1)/(ln (1/2))#
which is of the same form as in method 1. We would also get
Refer to the explanation.
The first ionization energy of an element is the energy needed to remove the outermost (valence) electron from a neutral atom in the gaseous state to form a cation.
The first ionization energy of an element has an inverse relationship to its atomic radius.
Across a period from left to right on the periodic table, first ionization energy increases and atomic radius decreases. Down a group, first ionization energy decreases and atomic radius increases.
The atoms of the alkali metals have larger radii than those of nonmetals. Because of this, the attraction of the positively charged atomic nucleus for the valence electron is much less than that of a nonmetal.
Therefore, the first ionization energy of an alkali metal is much less than that of a nonmetal. Because of this, alkali metals lose their single valence electron which produces a cation with a charge of
The atoms of nonmetals have smaller radii, therefore the positively charged nucleus has a greater attraction for the valence electrons.
So the first ionization energy for a nonmetal is much greater than that of an alkali metal, so nonmetals do not lose electrons in an ionic bond, but instead gain one or more electrons and form anions. An example is a neutral chlorine (Cl) atom, also in period 3, which has an atomic radius of
Refer to the diagram below.
The first ionization energy increases across a period.
#["H"^(+)] = 1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2))#
Note that this works best if
DISCLAIMER: DERIVATION!
Well, concentration is a state function, so we can simply choose acid 1 to go first, and acid 2 can go second, suppressed by the equilibrium of the first.
By writing out an ICE table, you would construct the mass action expression for acid 1 (say,
#"HA"(aq) rightleftharpoons "H"^(+)(aq) + "A"^(-)(aq)#
#K_(a1) = (["H"^(+)]_1["A"^(-)])/(["HA"]) = (["H"^(+)]_1alpha_1C_1)/((1 - alpha_1)C_1)#
Although we know that
Since we assume
#K_(a1) ~~ (["H"^(+)]_1alpha_1C_1)/(C_1) = alpha_1["H"^(+)]_1#
So,
#["H"^(+)]_1 ~~ K_(a1)/alpha_1#
The second acid, say
#"BH"^(+)(aq) rightleftharpoons "B"(aq) + "H"^(+)(aq)#
And the
#K_(a2) = (["B"]["H"^(+)])/(["BH"^(+)]) = (["H"^(+)]alpha_2C_2)/((1 - alpha_2)C_2)#
And since we also have
#K_(a2) ~~ alpha_2["H"^(+)]#
#["H"^(+)] ~~ K_(a2)/alpha_2#
with
#alpha_2C_2 = ["H"^(+)]_2 = K_(a2)/alpha_2 - alpha_1C_1# .
So, one form of this is
#["H"^(+)] = ["H"^(+)]_1 + ["H"^(+)]_2#
#= alpha_1C_1 + alpha_2C_2#
But we have built into
#color(red)(["H"^(+)] < (K_(a1))/alpha_1 + K_(a2)/alpha_2)#
It would be convenient to determine
#K_(a2)/alpha_2 = alpha_1C_1 + alpha_2C_2#
#0 = C_2alpha_2^2 + alpha_1C_1alpha_2 - K_(a2)#
This becomes a quadratic equation. If you wish to see it,
#color(green)(alpha_2) = (-(alpha_1C_1) pm sqrt(alpha_1^2C_1^2 - 4C_2(-K_(a2))))/(2C_2)#
#= color(green)((-(alpha_1C_1) pm sqrt(alpha_1^2C_1^2 + 4C_2K_(a2)))/(2C_2))#
And so,
#["H"^(+)] = alpha_1C_1 + (- (alpha_1C_1)/(2C_2) pm sqrt(alpha_1^2C_1^2 + 4C_2K_(a2))/(2C_2))C_2#
#= 1/2 alpha_1C_1 pm sqrt(((alpha_1C_1)/(2))^2 + K_(a2)C_2)#
And lastly, it would be convenient to know this in terms of
Since
#alpha_1 = sqrt(K_(a1)/C_1)#
Therefore:
#color(blue)(["H"^(+)]) = 1/2 sqrt(K_(a1)C_1) pm sqrt(((C_1)/(2))^2K_(a1)/C_1 + K_(a2)C_2)#
#= 1/2 sqrt(K_(a1)C_1) pm sqrt(1/4K_(a1)C_1 + K_(a2)C_2)#
#= color(blue)(1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2)))#
TESTING ON ACTUAL PROBLEM
And of course, we should try this on an actual problem.
Consider acetic acid (
#"HCHO"_2(aq) rightleftharpoons "H"^(+)(aq) + "CHO"_2^(-)(aq)#
#1.8 xx 10^(-4) = x^2/(0.50 - x) ~~ x^2/0.50#
So,
#["H"^(+)]_1 ~~ sqrt(1.8 xx 10^(-4) cdot 0.50) = "0.00949 M"#
and
#alpha_1 ~~ x/(["HCHO"_2]) = 0.019#
Let's see if we get the same
#1.8 xx 10^(-4) stackrel(?)(=) (alpha_1C_1)^2/((1 - alpha_1)C_1)#
#= (0.019 cdot 0.50)^2/((1 - 0.019)cdot0.50) ~~ 1.8 xx 10^(-4) color(blue)(sqrt"")#
Next, add in acetic acid.
#"HC"_2"H"_3"O"_2(aq) rightleftharpoons "H"^(+)(aq) + "C"_2"H"_3"O"_2^(-)(aq)#
#1.8 xx 10^(-5) = ((["H"^(+)]_1 + x)(x))/(2.00 - x) ~~ (["H"^(+)]_1 + x)x/2#
#~~ ["H"^(+)]_1x/2 + x^2/2#
The quadratic equation solves to be
#x = ["C"_2"H"_3"O"_2^(-)] = "0.00291 M"# ,which is also the
#"H"^(+)# contributed by the second acid.
And so, the total
#color(green)(["H"^(+)]) = "0.00949 M" + "0.00291 M" = color(green)ul("0.0124 M")#
This acid's percent dissociation at this concentration is
#alpha_2 = x/(["HC"_2"H"_3"O"_2]) = "0.00291 M"/"2.00 M" = 0.00145# .
So from the first form of the derived equation,
#color(green)(["H"^(+)] = alpha_1C_1 + alpha_2C_2)#
#= 0.019 cdot 0.50 + 0.00145 cdot 2.00 = color(green)ul("0.0124 M")# #color(blue)(sqrt"")#
Now to check the general formula we derived above.
#color(green)(["H"^(+)]) stackrel(?)(=) 1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2))#
#= 1/2 (sqrt(1.8 xx 10^(-4) cdot 0.50) pm sqrt(1.8 xx 10^(-4) cdot 0.50 + 4 cdot 1.8 xx 10^(-5) cdot 2.00))#
#= 1/2(sqrt(9.00 xx 10^(-5)) pm sqrt(9.00 xx 10^(-5) + 1.44 xx 10^(-4)))#
#=# #color(green)ul("0.0124 M")# #color(blue)(sqrt"")#
There shouldn't be, and that's why
Imagine that each
Then, take away the
What about
The first chemical bond made in a molecule is preferentially a
For
So, this particular orbital energy ordering takes place in the molecular orbital diagram, similar to the
where
#|E_(sigma_(2p_z))| > {|E_(pi_(2p_x))| = |E_(pi_(2p_y))|}# .
Since
#color(green)((sigma_(1s))^2(sigma_(1s^"*"))^2(sigma_(2s))^2(sigma_(2s^"*"))^2(pi_(2p_x))^1(pi_(2p_y))^1)#
Since
#:"B"-"B":#
That isn't a
Here's what I find.
What it does
Lunar caustic is an effective oxidizing agent for many organic compounds
It kills microorganisms and reacts with the compounds in human flesh.
Thus, it is used for removing warts, destroying damaged or diseased tissues, and for stopping superficial bleeding.
Origin of "caustic"
Caustic means "able to corrode organic tissue by chemical action".
This is just what lunar caustic does.
Improper use of lunar caustic can cause chemical burns.
Thus the origin of "caustic" in the name is understandable.
Origin of "lunar"
The alchemists believed that silver was associated with the moon.
They called silver luna (from Latin Luna = "the Moon"), so their name for silver nitrate was "lunar caustic".
Lithium:
Oxygen:
Nitrogen:
Potassium:
Lithium:
From its position, we know that it has
We also know that its
Putting it all together, we get
Oxygen:
From its position in the periodic table, we know that it has
We also know that:
Putting it all together, we get
Nitrogen:
Nitrogen is directly to the left of oxygen in the periodic table. This tells us that it has one less electron than oxygenâ€”therefore, its electron configuration is the exact same as oxygen's, except with one less electron in the valence energy level.
Oxygen's electron configuration is
After taking one electron from that, it becomes
Potassium:
From its position in the periodic table, we know that it has
We also know that:
Its
Its
Its
Its
Its
Putting it all together, we get
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