ALEKS 360 AC INTRD CHEM >I<
5th Edition
ISBN: 9781260977585
Author: BAUER
Publisher: MCG
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Chapter 15, Problem 101QP
Interpretation Introduction
Interpretation: The amount of polonium-210 left after a period of
Concept Introduction:
The half-life of a nuclide refers to the time period after which half the nuclide would have decayed. The half-life of a nuclide is always the same, irrespective of the amount of nuclide, temperature, or any other conditions. After one half-life, only one half of the given nuclide present at the beginning of the time period remains.
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NMR spectrum of ethyl acetate has signals whose chemical shifts are indicated below. Which hydrogen or set of hydrogens corresponds to the signal at
4.1 ppm? Select the single best answer.
The
H
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2
A
ethyl acetate
H NMR: 1.3 ppm, 2.0 ppm, 4.1 ppm
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OC
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B
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How many signals do you expect in the H NMR spectrum for this molecule?
Br Br
Write the answer below.
Also, in each of the drawing areas below is a copy of the molecule, with Hs shown. In each copy, one of the H atoms is colored red. Highlight in red all other H
atoms that would contribute to the same signal as the H already highlighted red
Note for advanced students: In this question, any multiplet is counted as one signal.
1
Number of signals in the 'H NMR spectrum.
For the molecule in the top drawing area, highlight in red any other H atoms that will contribute to
the same signal as the H atom already highlighted red.
If no other H atoms will contribute, check the box at right.
Check
For the molecule in the bottom drawing area, highlight in red any other H atoms that will contribute
to the same signal as the H atom already highlighted red.
If no other H atoms will contribute, check the box at right.
O
✓
No additional Hs to color in top
molecule
ง
No additional Hs to color in bottom…
in the kinetics experiment, what were the values calculated? Select all that apply.a) equilibrium constantb) pHc) order of reactiond) rate contstant
Chapter 15 Solutions
ALEKS 360 AC INTRD CHEM >I<
Ch. 15 - Prob. 1QCCh. 15 - Prob. 2QCCh. 15 - Prob. 3QCCh. 15 - Prob. 4QCCh. 15 - Prob. 5QCCh. 15 - Prob. 6QCCh. 15 - Prob. 1PPCh. 15 - Prob. 2PPCh. 15 - Prob. 3PPCh. 15 - Prob. 4PP
Ch. 15 - Prob. 5PPCh. 15 - Prob. 6PPCh. 15 - Prob. 7PPCh. 15 - Prob. 8PPCh. 15 - Prob. 9PPCh. 15 - Prob. 10PPCh. 15 - Prob. 11PPCh. 15 - Prob. 1QPCh. 15 - Prob. 2QPCh. 15 - Prob. 3QPCh. 15 - Prob. 4QPCh. 15 - Prob. 5QPCh. 15 - Prob. 6QPCh. 15 - Prob. 7QPCh. 15 - Prob. 8QPCh. 15 - Prob. 9QPCh. 15 - Prob. 10QPCh. 15 - Prob. 11QPCh. 15 - Prob. 12QPCh. 15 - Prob. 13QPCh. 15 - Prob. 14QPCh. 15 - Prob. 15QPCh. 15 - Prob. 16QPCh. 15 - Prob. 17QPCh. 15 - Prob. 18QPCh. 15 - Prob. 19QPCh. 15 - Prob. 20QPCh. 15 - Prob. 21QPCh. 15 - Prob. 22QPCh. 15 - Prob. 23QPCh. 15 - Prob. 24QPCh. 15 - Prob. 25QPCh. 15 - Prob. 26QPCh. 15 - Prob. 27QPCh. 15 - Prob. 28QPCh. 15 - Prob. 29QPCh. 15 - Prob. 30QPCh. 15 - Prob. 31QPCh. 15 - Prob. 32QPCh. 15 - Prob. 33QPCh. 15 - Prob. 34QPCh. 15 - Prob. 35QPCh. 15 - Prob. 36QPCh. 15 - Prob. 37QPCh. 15 - Prob. 38QPCh. 15 - Prob. 39QPCh. 15 - Prob. 40QPCh. 15 - Prob. 41QPCh. 15 - Prob. 42QPCh. 15 - Prob. 43QPCh. 15 - Prob. 44QPCh. 15 - Prob. 45QPCh. 15 - Prob. 46QPCh. 15 - Prob. 47QPCh. 15 - Prob. 48QPCh. 15 - Prob. 49QPCh. 15 - Prob. 50QPCh. 15 - Prob. 51QPCh. 15 - Prob. 52QPCh. 15 - Prob. 53QPCh. 15 - Prob. 54QPCh. 15 - Prob. 55QPCh. 15 - Prob. 56QPCh. 15 - Prob. 57QPCh. 15 - Prob. 58QPCh. 15 - Prob. 59QPCh. 15 - Prob. 60QPCh. 15 - Prob. 61QPCh. 15 - Prob. 62QPCh. 15 - Prob. 63QPCh. 15 - Prob. 64QPCh. 15 - Prob. 65QPCh. 15 - Prob. 66QPCh. 15 - Prob. 67QPCh. 15 - Prob. 68QPCh. 15 - Prob. 69QPCh. 15 - Prob. 70QPCh. 15 - Prob. 73QPCh. 15 - Prob. 74QPCh. 15 - Prob. 75QPCh. 15 - Prob. 76QPCh. 15 - Prob. 77QPCh. 15 - Prob. 78QPCh. 15 - Prob. 79QPCh. 15 - Prob. 80QPCh. 15 - Prob. 81QPCh. 15 - Prob. 82QPCh. 15 - Prob. 83QPCh. 15 - Prob. 84QPCh. 15 - Prob. 85QPCh. 15 - Prob. 86QPCh. 15 - Prob. 87QPCh. 15 - Prob. 88QPCh. 15 - Prob. 89QPCh. 15 - Prob. 90QPCh. 15 - Prob. 91QPCh. 15 - Prob. 92QPCh. 15 - Prob. 93QPCh. 15 - Prob. 94QPCh. 15 - Prob. 95QPCh. 15 - Prob. 96QPCh. 15 - Prob. 97QPCh. 15 - Prob. 98QPCh. 15 - Prob. 99QPCh. 15 - Prob. 100QPCh. 15 - Prob. 101QPCh. 15 - Prob. 102QPCh. 15 - Prob. 103QPCh. 15 - Prob. 104QPCh. 15 - Prob. 105QPCh. 15 - Prob. 106QPCh. 15 - Prob. 107QPCh. 15 - Prob. 108QPCh. 15 - Prob. 109QPCh. 15 - Prob. 110QPCh. 15 - Prob. 111QPCh. 15 - Prob. 112QPCh. 15 - Prob. 113QPCh. 15 - Prob. 114QPCh. 15 - Prob. 115QPCh. 15 - Prob. 116QPCh. 15 - Prob. 117QPCh. 15 - Prob. 118QPCh. 15 - Prob. 119QPCh. 15 - Prob. 120QPCh. 15 - Prob. 121QPCh. 15 - Prob. 122QPCh. 15 - Prob. 123QP
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- true or false, given that a 20.00 mL sample of NaOH took 24.15 mL of 0.141 M HCI to reach the endpoint in a titration, the concentration of the NaOH is 1.17 M.arrow_forwardin the bromothymol blue experiment, pKa was measured. A closely related compound has a Ka of 2.10 x 10-5. What is the pKa?a) 7.1b) 4.7c) 2.0arrow_forwardcalculate the equilibrium concentration of H2 given that K= 0.017 at a constant temperature for this reaction. The inital concentration of HBr is 0.050 M.2HBr(g) ↔ H2(g) + Br2(g)a) 4.48 x 10-2 M b) 5.17 x 10-3 Mc) 1.03 x 10-2 Md) 1.70 x 10-2 Marrow_forward
- true or falsegiven these two equilibria with their equilibrium constants:H2(g) + CI2(l) ↔ 2HCI(g) K= 0.006 CI2(l) ↔ CI2(g) K= 0.30The equilibrium contstant for the following reaction is 1.8H2(g) + CI2 ↔ 2HCI(g)arrow_forwardI2(g) + CI2(g) ↔ 2ICIK for this reaction is 81.9. Find the equilibrium concentration of I2 if the inital concentration of I2 and CI2 are 0.010 Marrow_forwardtrue or false,the equilibrium constant for this reaction is 0.50.PCI5(g) ↔ PCI3(g) + CI2(g)Based on the above, the equilibrium constant for the following reaction is 0.25.2PCI5(g) ↔. 2PCI3(g) + 2CI2(g)arrow_forward
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