1st round 2011 problems v5 - Ķīmijas didaktikas centrs · 2010-09-28 · 1 2011 1st round,...
TRANSCRIPT
1
www.biosan.lv
2011 1ST ROUND, PROBLEMS
Solve all or some of the problems given on the next pages and write your full solutions in MS Word, Excel documents or *.pdf files. It is also acceptable if you scan your material and insert as picture in mentioned file formats.
If there are some explanations required, write them in English. Please, send your answers to: [email protected] till 08.11.2010. at 3:00 (Latvian time; +2). Answers sent after this deadline will not be graded.
File name must consist from your name, last name (in English) and country, for example, John_Black_England.doc (or *.docx etc.). If you do not name the file as described you will receive 3 point penalty. For all the correctly solved problems you can get the maximum of 30 points. The exact amount of points for each task is given at the top of each problem.
All the students who will be taking part in at least one round will participate in the final round which will be held on the web on February 27th. During the final round you will have to solve the multiple‐choice test. More information can be found in BCC regulations. Competition organizers and problem authors:
Kaspars Veldre Vladislav Ivaništšev Egle Maksimaviciute Filip Topić
PhD student, University of Latvia, Department of Chemistry
PhD student, Tartu University,
Institute of Chemistry
2nd year BSc student, Vilnius University, Department of
Chemistry
Croatia
Peter Holzhauser Agris Bērziņš Karina Kizjakina Alons Lends
Czech Republic MS student, University of Latvia, Department
of Chemistry
PhD student, Virginia Polytechnic Institute
and State University, Department of Chemistry
BSc student, Riga Technical University
Good luck with problem solving! ☺
If you have any questions, you can address them to us in English by sending them to: [email protected] Feel free to ask!
Proble
Sulfur
A 5.00 L cloand O2 (30%container w
SO2 +0.5 O2
1. Whafte
1.00 L of waclosed.
2. Wrisolu
3. Whadd
4. Wh(O2)
5. Is itlowSho
All requiredof informat
Proble
Bless y
If too many Scheme 1
em 1 (Li
contai
sed containe% of total volwas kept at 2
SO3 rea
at will be theer the equilib
ater was add
ite the reactiution? hat would be dition of the ghat would be ) of the totalt possible chawer than in paow the calcul
SO2 SO3 O2
d constants cion.
em 2 (Es
you! (10
y drugs are p
ithuani
ining ga
er was filled lume) gases,5°C.
action occurs
e final volumbrium is reac
ded into the s
ion that occu
the pH of thgases??) wothe pH of thl volume (at ange concenart 2 by chanlations.
can be found
stonia)
points)
rovided agai
ia)
ases (8 p
with SO2 (70, the temper
s at 25°C.
me of the O2 iched?
same contai
urs upon the
he solution if ould be at 200he solution if 25°C)? tration of hynging temper
ΔfHo/ (kJ
‐296.83 ‐395.72 0
online or in
)
nst one illne
2
points)
0% of total voature inside
in that conta
ner keeping
addition of
f the tempera0°C? After adf the initial pr
ydrogen ionsrature or init
mol‐1)
handbooks.
ess, then it m
olume) the
ainer
it still
water? Wha
ature inside tdding water roportion of
s in solution ttial proportio
Smo/ (J
248.22256.76205.13
. References
means that th
at is the pH of
the containethe solutiongases is 30 %
to be two timon of gases?
mol‐1 K‐1) 2 6 38
should be g
his illness can Ant
f the resultin
er (during the is cooled to % (SO2) and 7
mes higher or? If so then ho
iven to the s
nnot be healton Chekhov
ng
e 25°C. 70 %
r ow?
source
ed. v
Write the st
Do you agreof its lone e
Scheme 2
Write the st
Write downstereocente
Proble
Quantu
β‐Cred‐orange the most ccontains a cThe π electconsidered atom contri
tructures for
ee that in caselectron pair
tructures for
n the structuers.
em 3 (La
um org
Carotene is apigment, abcommon forconjugated crons in this to be particlibutes to one
r compounds
se when nitrthat plays th
r compounds
res for all of
atvia)
ganic ch
an organic cobundant in prm of carotechain of 22 cconjugated cles in a one de π electron.
s A‐E.
rogen atom che role of the
s A‐E, Х.
the compou
hemistr
ompound thplants and frene. It is a arbon atomschain are dedimensional .
3
carries threee forth subst
und B stereo
ry (4 po
at is classifieruits. As a caprecursor (s with an aveelocalized anbox of 3.17
different sutituent? Why
isomers and
oints)
ed as a terparotene withinactive forerage internund, to a rougnm (that is 2
ubstituents ity?
assign confi
enoid. It is ah beta‐rings m) of vitamuclear separgh approxima22x14 pm)le
t is chiral, be
igurations at
a strongly‐coat both end
min A. β‐Carration of 144ation, they cngth. Each c
cause
t their
olored s, it is rotene 4 ppm. can be arbon
p
1.
2.
Proble
Chemispoints)
Part 1 – Rad
Radioactive
two groups
Y X 4-A2Z
AZ → −
Y X A1Z
AZ → +
Y X A1Z
AZ −→
Complete fo
...... n10 →
.... Be84 →
N C 147
146 →
..... K 4019 →
Ga 683
6831 →
Part 2 – Me
For measurbequerel cominute), Bqvolume of radioactivityof radium‐2
1) Calculate
Later, half‐lbut measur
2) Calculate
Calculate th(HOMO) anCalculate thbetween th
em 4 (Es
stry of )
dioactive de
e decays are
: α‐ and β‐de
He Y 42+
e Y 01-+
e Y 01++
ollowing equ
e 01-+
He .. 42+
...... N +
e .. 01-+
...... Y 830 +
easuring radi
ring activity oorresponds q/kg or Bq/mradioactive y. One curie 226.
e half‐life of 2
ife of radiumrement unit C
e mass or rad
he differenced the lowesthe wavelengese two stat
stonia)
f radioa
cay equatio
classified by
ecays:
α‐decay,
β‐‐decay,
β+‐decay.
uations and,
T9943
11348
10953
14460
14862
ioactivity
of radioactivto one dps
mL are derivematerial. Beis equal to 3
226Ra (in year
m‐226 was dCi was not ch
dium which h
e in energy bt unoccupiedgth of radiatites.
active p
ns
y rules of Faj
if it is possib
...... Tc +→
...... Cd→
...... I +→
...... Nd→
N Sm 14460→
ve particles Ss (decay pered from Bq tesides SI uni3.70×1010 Bq
rs) using def
determined mhanged and
has activity o
4
between thed level (LUMOion that wou
particle
jans and Sod
ble, determin
...... +
e 01-+
e01-
He 42+
...... Nd +
SI measuremr second). Mthat way to its non‐Si unq and origina
finition of cur
more preciseit is still bein
of 1 Ci.
e highest occO). uld be absor
es (8
ddy into
ne the type o
Pb20482
Tl20681 →
Bi21083 →
Rn22286
Ra22688
ment units, bMeasuremenshow the nnit, curie (Cially it was de
rium.
ely and it appng used for a
cupied π elec
rbed in prod
of radioactive
...... 42+→
.... ...... +→
.. Tl 20681 +→
Po 21884 +→
... ...... +→
ecquerels (Bnt units as number of dei), is also usfined as acti
peared to bepproximate
ctron energy
ducing a tran
e decay.
He
..
....
...... +
....
Bq), are useddpm (decayecays per msed for descvity of 1.00 g
e T½ = 1620 calculations.
y level
nsition
d. One ys per ass or cribing grams
years, .
5
Radium was discovered in 1902 by Pierre Curie and Marie Sklodowska‐Curie. From 8000 tons of uranium ore waste they obtained 100 mg of radium chloride.
3) Calculate activity of obtained radium chloride in Bq per kilogram (Bq/kg).
In 1907 Marie Curie gifted 1.00 g of radium to Institute of Radium in Paris (now Institute of Curie), where it is still kept nowadays.
4) Calculate amount of radium (in moles) which was not decomposed in the time period from 1907 to2010.
5) Calculate mass of radium (in g) decomposed in this time period.
6) Calculate in what year practically all radium will be decomposed (99.99%).
Part 3 – Equipment for measuring radioactivity
Radioactivity of particles is measured by special equipment. Not all measuring devices are up‐to‐date so it is very hard to determine the correct value of radioactivity. With the aim to improve radioactivity measurement, instruments are calibrated using specially prepared specimens.
Such material is S3516 (T½ = 87.9 days) with initial radioactivity 0.0100 μCi. In 200 days, radioactivity of
this material, determined by Geiger counter, decreases to 2600 dcm.
1) Calculate actual radioactivity that S3516 will have in 200 days (in Bq).
2) Calculate sensitivity of Geiger counter (in %).
3) Calculate radioactivity value determined by this Geiger counter after 7 days for 100 mg 51‐Cr sample (T½ = 27.8 days).
4) Calculate mass fraction of 51‐Cr in obtained sample (after 7 days).
Part 4 – Dangerousness of radioactive elements
At the moment of Earth formation, radioactive elements were part of several chemical compounds. Some scientists associate high temperatures in Earth core with the radioactive heat. Existence of radioactive elements with long half‐lifes (for example: uranium‐238 – 4.5.109 years; thorium‐232 – 1.4.1010 years; potassium‐40 – 1.28.109 years; rubidium‐87 – 6.75.1010 etc.) is supporting this statement. Other elements with shorter half‐lifes are formed at decay of uranium‐238, thorium‐232, and uranium‐235. Radium, polonium, and radon are among those newly formed elements.
Due to effect of different waves from universe on the Earth atmosphere, radioactive carbon 14‐C (T½ = 5720 years) and tritium 3‐H (T½ = 12.4 years) are formed. Radioactive background exists at all times, and is formed from both natural and man‐made radionuclides. For example, contents of radioactive isotope of potassium 40‐K (T½ = 1.275 ∙10
6 years) in nature is equal to 0.0117%. All other isotopes of potassium are stable.
1) Calculate radioactivity of metallic potassium (m = 20 g), found in school laboratory.
2) State at least two reasons why potassium from school laboratory is considered to be safe.
Especially dangerous isotope is strontium‐90, which replaces calcium in live organisms and accumulates in bones.
3) Explain why strontium‐90 is so dangerous, name two illnesses caused by radioactive strontium.
As a fact, cigarette contains from 3 to 24 mBq of 210‐Po. After entering human organism, most part of 210‐Po accumulates in bones, teeth, and also in kidneys and liver.
4) Calculate time that is required for complete (99.99%) decomposition of 210‐Po (T½ = 138.4 days) after smoking one cigarette.