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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.

  • After one half-lives, #[A] = 1/2[A]_0#.
  • After two half-lives, #[A] = 1/4[A]_0#.
  • After three half-lives, #[A] = 1/8[A]_0#.
  • After four half-lives, #[A] = 1/16[A]_0#.

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, #n#, is simply the total time divided by the known half-life.

Thus, #n = t//t_"1/2"#, and we have the current concentration as a function of the starting concentration, half-life, and total time passed:

#barul|stackrel(" ")(" "[A] = (1/2)^(t//t_"1/2") [A]_0" ")|#

Given that #10%# of the sample remains, we know that

#[A] = 0.1[A]_0#,

and thus,

#(1/2)^(t//t_"1/2") = 0.1#.

To solve for #t#, take the #ln# of both sides.

#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 #10%# of the sample to remain:

#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 #[A] = 0.1[A]_0# as mentioned earlier.

#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 #"145.5 s"# in this way.

Answer:

Refer to the explanation.

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.

#X_"(g)" + "energy"##rarr##"X"_"(g)"^(+) + e^(-)"#

The first ionization energy of an element has an inverse relationship to its atomic radius.

http://www.bianoti.com/ionization-energy.html

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.

http://imgarcade.com/atomic-radius-example.html

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 #1^+#. An example is the neutral sodium (Na) atom, in period 3, which has an atomic radius of #"190 picometers"# #("pm")#, and a first ionization energy of #"496 kilojoules/mol"# #("kJ/mol")#.

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 #"79 pm"#, and a first ionization energy of #"1251 kJ/mol"#. Compare these values with a neutral Na atom previously mentioned.

Refer to the diagram below.

http://www.vias.org/genchem/atomstruct_12433_05.html

The first ionization energy increases across a period. #"H"# (not labeled) to #"He"# is period 1, #"Li"# to #"Ne"# is period 2, #"Na"# to #"Ar"# is period 3, and so on.

#["H"^(+)] = 1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2))#

Note that this works best if #K_(a1)# and #K_(a2)# are both on the order of #10^(-5)# or less, i.e. the acids are sufficiently weak.


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"#):

#"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 #["H"^(+)]_1 = ["A"^(-)]# in this first process, we choose to write them distinct from each other.

Since we assume #alpha_1# #"<<"# #1#, i.e. #K_(a1) < 10^(-5)# or so, then rewrite this as...

#K_(a1) ~~ (["H"^(+)]_1alpha_1C_1)/(C_1) = alpha_1["H"^(+)]_1#

So,

#["H"^(+)]_1 ~~ K_(a1)/alpha_1#

The second acid, say #"BH"^(+)#, now acts.

#"BH"^(+)(aq) rightleftharpoons "B"(aq) + "H"^(+)(aq)#

And the #["H"^(+)]_1# from acid 1 now is the initial concentration for the #"BH"^(+)# dissociation.

#K_(a2) = (["B"]["H"^(+)])/(["BH"^(+)]) = (["H"^(+)]alpha_2C_2)/((1 - alpha_2)C_2)#

And since we also have #alpha_2# #"<<"# #1#,

#K_(a2) ~~ alpha_2["H"^(+)]#

#["H"^(+)] ~~ K_(a2)/alpha_2#

with #["H"^(+)]# being the net #["H"^(+)]# concentration. By subtracting out the contribution from acid 1,

#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 #alpha_2# the suppressed equilibrium, and so,

#color(red)(["H"^(+)] < (K_(a1))/alpha_1 + K_(a2)/alpha_2)#

It would be convenient to determine #alpha_2# in terms of #alpha_1#, but it won't look nice at first.

#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 #K_(a1)# instead of #alpha_1#, since that requires information we may not already have.

Since #K_(a1) ~~ alpha_1["H"^(+)]_1 ~~ alpha_1^2C_1#, we can say that

#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 (#K_a = 1.8 xx 10^(-5)#) and formic acid (#K_a = 1.8 xx 10^(-4)#), for #C_1 = "0.50 M"# and #C_2 = "2.00 M"#. Suppose we let formic acid dissociate first.

#"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 #K_a#.

#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 #"H"^(+)# is:

#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 #"B"_2# is very unstable; it has two orthogonally-localized half-#pi# bonds, which is quite a weak bond.

Imagine that each #"B"# takes the place of a #"CH"# on #"HC"-="CH"#.

https://chem.libretexts.org/

Then, take away the #sigma# bond, the two #"H"# atoms, and one electron from each of the two full #pi# bonds, and you have a structure similar to #"B"_2#.

What about #"Li"_2# "molecule"? Is that stable? (Its bond length is #"267.3 pm"#, over twice the length of an average bond.)


The first chemical bond made in a molecule is preferentially a #bbsigma# bond.

#sigma# bonds are formed from a direct atomic orbital overlap. In comparison, #pi# bonds are sidelong overlaps and thus, #sigma# overlaps are made preferentially because they form the stronger bond.

#"B"_2# contains two boron atoms, which each use a basis of a #1s#, a #2s#, and three #2p# atomic orbitals.

For #"Li"_2# through #"N"_2#, there exists an orbital mixing effect that makes the #sigma# molecular orbital for a #2p_z-2p_z# overlap (#sigma_(g(2p))#) higher in energy than the #pi# molecular orbitals for a #2p_(x//y)-2p_(x//y)# overlaps (#pi_(u(2p))#).

Inorganic Chemistry, Miessler et al., Ch. 5.2.3

So, this particular orbital energy ordering takes place in the molecular orbital diagram, similar to the #"C"_2# molecule:

where #|E_(sigma_(2p_z))| > {|E_(pi_(2p_x))| = |E_(pi_(2p_y))|}#.

Since #"C"# has one more electron, #"B"_2# would have a similar MO diagram, EXCEPT for two fewer electrons. This, at first glance, seems to suggest that #"B"_2# has two half-#pi# bonds, with a molecular electron configuration of:

#color(green)((sigma_(1s))^2(sigma_(1s^"*"))^2(sigma_(2s))^2(sigma_(2s^"*"))^2(pi_(2p_x))^1(pi_(2p_y))^1)#

Since #"B"_2# is paramagnetic with two electrons, in order to make that bond, there are indeed two half-#bbpi# bonds, which form what we represent improperly in line notation as a #sigma# bond... and it isn't actually a #sigma# bond.

#:"B"-"B":#

That isn't a #sigma# bond, but two half-#pi# bonds. We would then expect #"B"_2# to be very unstable, which it is. (The lone pairs are from the #sigma_(2s)# and #sigma_(2s)^"*"#, since bonding + antibonding filled = nonbonding.)

Answer:

Here's what I find.

Explanation:

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".

Answer:

Lithium: #1s^2 2s^1#
Oxygen: #1s^2 2s^2 2p^4#
Nitrogen: #1s^2 2s^2 2p^3#
Potassium: #1s^2 2s^2 2p^6 3s^2 3p^6 4s^1#

Explanation:

Lithium:

PTable

From its position, we know that it has #1# valence electron in the #2s# orbital series (because it's in the second period): #2s^1#.

We also know that its #1s# orbital is full, because to get to lithium in the periodic table, we have to pass #1s#. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #1s# orbital: #1s^2#.

Putting it all together, we get #1s^2 2s^1#.

Oxygen:

PTable

From its position in the periodic table, we know that it has #4# valence electrons in the #2p# orbital series (because it's in the second period): #2p^4#.
We also know that:

  • Its #1s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #1s# orbital: #1s^2#.
  • Its #2s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #2s# orbital: #2s^2#.

Putting it all together, we get #1s^2 2s^2 2p^4#.

Nitrogen:

PTable

Nitrogen is directly to the left of oxygen in the periodic table. This tells us that it has one less electron than oxygen—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 #1s^2 2s^2 2p^4#.
After taking one electron from that, it becomes #1s^2 2s^2 2p^3#—nitrogen.

Potassium:

PTable

From its position in the periodic table, we know that it has #1# valence electron in the #4s# orbital series (because it's in the #s# block of the fourth period): #4s^1#.
We also know that:

  • Its #1s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #1s# orbital: #1s^2#.

  • Its #2s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #2s# orbital: #2s^2#.

  • Its #2p# orbital is full. There are #6# electrons in an #s# orbital; this means that it has #6# electrons in its #2p# orbital: #2p^6#.

  • Its #3s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #3s# orbital: #3s^2#.

  • Its #3p# orbital is full. There are #6# electrons in an #s# orbital; this means that it has #6# electrons in its #3p# orbital: #3p^6#.

Putting it all together, we get #1s^2 2s^2 2p^6 3s^2 3p^6 4s^1#.

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