Introduction to Chemistry
4th Edition
ISBN: 9781259288722
Author: BAUER
Publisher: MCG
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Chapter 12, Problem 136QP
Interpretation Introduction
Interpretation:
In the given two-step mechanism, the intermediate is to be identified and the presence of a catalyst is to be determined. The catalyst substance is also to be named.
Concept Introduction:
Catalyst is a chemical substance used to increase or decrease the
In a
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LABORATORY REPORT FORM
Part I. Determination of the Formula of a Known Hydrate
1. Mass of empty evaporating dish
3. Mass of hydrate
Using subtraction
or mass by difference,
find the mass of
the hydrate
76.96
-75.40
75.40g
76.968
1.568
01.56
76.90 g
2. Mass of evaporating dish + hydrate
4. Mass of evaporating dish + hydrate (after heating)
First 76.98 g
Third 76.66g
Second
Fourth (if necessary) 76.60g
5. Mass of anhydrate
6. Mass of water lost by the hydrate
7. Percent of water of hydration
(Show Calculations)
8. Moles of water
(Show Calculations)
mol
mass of water
=
MM of water (g/m)
9. Moles of anhydrate
(Show Calculations)
10. Ratio of moles of water to moles of anhydrate
11 F(Show Calculations)
11. Formula of hydrate
- Mass of water (g) x 100
% water hydration
g
g
%
Mass of hydrate (9) x IC
% = (Mass of hydrate- mass of an)
mass of hydrate (g)
% = (1.569-
× 100=
mol
1.569
mol Mol Mass of
anhydrate/MM
of anhydrate
12. What was the color of the hydrate?
blue
What was the color of the…
compared t-critical with t-calculated and 95% confidence interval to answer this question
Comparing two means. Horvat and co-workers used atomic absorption spectroscopy to determine the concentration of Hg in coal fly ash. Of particular interest to the authors was developing an appropriate procedure for digesting samples and releasing the Hg for analysis. As part of their study they tested several reagents for digesting samples. Their results using HNO3 and using a 1+3 mixture of HNO3 and HCl are shown here. All concentrations are given as ppb Hg sample.
HNO3: 161, 165, 160, 167, 166
1+3 HNO3–HCl: 159, 145, 140, 147, 143, 156
Determine whether there is a significant difference between these methods at the 95% confidence interval.
Chapter 12 Solutions
Introduction to Chemistry
Ch. 12 - Prob. 1QCCh. 12 - Prob. 2QCCh. 12 - Prob. 3QCCh. 12 - Prob. 4QCCh. 12 - Prob. 5QCCh. 12 - Prob. 6QCCh. 12 - Prob. 1PPCh. 12 - Prob. 2PPCh. 12 - Prob. 3PPCh. 12 - Prob. 4PP
Ch. 12 - Prob. 5PPCh. 12 - Prob. 6PPCh. 12 - Prob. 7PPCh. 12 - Prob. 8PPCh. 12 - Prob. 9PPCh. 12 - Prob. 10PPCh. 12 - Consider the following equilibrium:...Ch. 12 - Prob. 12PPCh. 12 - Prob. 1QPCh. 12 - Match the key terms with the descriptions...Ch. 12 - Prob. 3QPCh. 12 - Prob. 4QPCh. 12 - Prob. 5QPCh. 12 - Prob. 6QPCh. 12 - Prob. 7QPCh. 12 - Prob. 8QPCh. 12 - Prob. 9QPCh. 12 - Prob. 10QPCh. 12 - Prob. 11QPCh. 12 - Prob. 12QPCh. 12 - Prob. 13QPCh. 12 - Prob. 14QPCh. 12 - Prob. 15QPCh. 12 - Prob. 16QPCh. 12 - Prob. 17QPCh. 12 - Prob. 18QPCh. 12 - Prob. 19QPCh. 12 - Prob. 20QPCh. 12 - Prob. 21QPCh. 12 - Prob. 22QPCh. 12 - Prob. 23QPCh. 12 - Prob. 24QPCh. 12 - Prob. 25QPCh. 12 - Prob. 26QPCh. 12 - Prob. 27QPCh. 12 - Prob. 28QPCh. 12 - Prob. 29QPCh. 12 - Prob. 30QPCh. 12 - Prob. 31QPCh. 12 - Prob. 32QPCh. 12 - Prob. 33QPCh. 12 - Prob. 34QPCh. 12 - Prob. 35QPCh. 12 - Prob. 36QPCh. 12 - Prob. 37QPCh. 12 - Prob. 38QPCh. 12 - Prob. 39QPCh. 12 - Prob. 40QPCh. 12 - Prob. 41QPCh. 12 - Prob. 42QPCh. 12 - Prob. 43QPCh. 12 - Prob. 44QPCh. 12 - Prob. 45QPCh. 12 - Prob. 46QPCh. 12 - Prob. 47QPCh. 12 - Prob. 48QPCh. 12 - Prob. 49QPCh. 12 - Prob. 50QPCh. 12 - Prob. 51QPCh. 12 - Prob. 52QPCh. 12 - Prob. 53QPCh. 12 - Prob. 54QPCh. 12 - Prob. 55QPCh. 12 - Prob. 56QPCh. 12 - Prob. 57QPCh. 12 - Prob. 58QPCh. 12 - Prob. 59QPCh. 12 - Prob. 60QPCh. 12 - Prob. 61QPCh. 12 - Prob. 62QPCh. 12 - Prob. 63QPCh. 12 - Prob. 64QPCh. 12 - Prob. 65QPCh. 12 - Prob. 66QPCh. 12 - Prob. 67QPCh. 12 - Prob. 68QPCh. 12 - Prob. 69QPCh. 12 - Prob. 70QPCh. 12 - Prob. 71QPCh. 12 - Prob. 72QPCh. 12 - Prob. 73QPCh. 12 - Prob. 74QPCh. 12 - Prob. 75QPCh. 12 - Prob. 76QPCh. 12 - Prob. 77QPCh. 12 - Prob. 78QPCh. 12 - Prob. 79QPCh. 12 - Prob. 80QPCh. 12 - Prob. 81QPCh. 12 - Prob. 82QPCh. 12 - Prob. 83QPCh. 12 - Prob. 84QPCh. 12 - Prob. 85QPCh. 12 - Prob. 86QPCh. 12 - Prob. 87QPCh. 12 - Prob. 88QPCh. 12 - Prob. 89QPCh. 12 - Prob. 90QPCh. 12 - Prob. 91QPCh. 12 - Prob. 92QPCh. 12 - Prob. 93QPCh. 12 - Prob. 94QPCh. 12 - Prob. 95QPCh. 12 - Prob. 96QPCh. 12 - Prob. 97QPCh. 12 - Prob. 98QPCh. 12 - Prob. 99QPCh. 12 - Prob. 100QPCh. 12 - Prob. 101QPCh. 12 - Prob. 102QPCh. 12 - Prob. 103QPCh. 12 - Prob. 104QPCh. 12 - Prob. 105QPCh. 12 - Prob. 106QPCh. 12 - Prob. 107QPCh. 12 - Prob. 108QPCh. 12 - Prob. 109QPCh. 12 - Prob. 110QPCh. 12 - Prob. 111QPCh. 12 - Prob. 112QPCh. 12 - Prob. 113QPCh. 12 - Prob. 114QPCh. 12 - Prob. 115QPCh. 12 - Prob. 116QPCh. 12 - Prob. 117QPCh. 12 - Prob. 118QPCh. 12 - Prob. 119QPCh. 12 - Prob. 120QPCh. 12 - Prob. 121QPCh. 12 - Prob. 122QPCh. 12 - Prob. 123QPCh. 12 - Prob. 124QPCh. 12 - Prob. 125QPCh. 12 - Prob. 126QPCh. 12 - Prob. 127QPCh. 12 - Prob. 128QPCh. 12 - Prob. 129QPCh. 12 - Prob. 130QPCh. 12 - Prob. 131QPCh. 12 - Prob. 132QPCh. 12 - Prob. 133QPCh. 12 - Prob. 134QPCh. 12 - Prob. 135QPCh. 12 - Prob. 136QPCh. 12 - Prob. 137QPCh. 12 - Prob. 138QPCh. 12 - Prob. 139QPCh. 12 - Prob. 140QPCh. 12 - Prob. 141QPCh. 12 - Prob. 142QPCh. 12 - Prob. 143QP
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- Comparison of experimental data to “known” value. Monna and co-workers used radioactive isotopes to date sediments from lakes and estuaries.21 To verify this method they analyzed a 208Po standard known to have an activity of 77.5 decays/min, obtaining the following results. 77.09, 75.37, 72.42, 76.84, 77.84, 76.69, 78.03, 74.96, 77.54, 76.09, 81.12, 75.75 Do the results differ from the expected results at the 95% confidence interval?arrow_forwardExplain the difference between the propagated uncertainty and the standard deviation. Which number would you use to describe the uncertainty in the measurement? if the standard deviation is 0.01 and the propagated uncertainty is 0.03arrow_forwardPropagation of uncertainty. Find the absolute and percent relative uncertainty assuming the ±-values are random error. 7.65±0.04 + 5.28±0.02 – 1.12±0.01 85.6±0.9 × 50.2±0.7 ÷ 13.8±0.5 [4.88±0.07 + 3.22±0.05] / 1.53±0.02arrow_forward
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