The differences in melting and boiling points of the given compounds have to be accounted. Concept Introduction: Intermolecular forces in a molecule are the important factor which is responsible for the physical changes of the molecule. The intermolecular forces in gases are negligibly small and the forces are comparatively stronger in liquids and the intermolecular forces are highest in solids. The stronger the intermolecular forces, closer will be the molecules in contact and thus higher will be the boiling point and melting point of the substance. Hydrogen bonding is a special type of dipole-dipole interaction in a polar bond which has hydrogen atom and a highly electronegative atom such as Nitrogen, Oxygen and Fluorine. Hydrogen bonding can also be defined as coulombic attraction between the hydrogen atom and an electronegative atom.

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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…

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Chapter 18, Problem 18.14P

Interpretation Introduction

Interpretation:

The differences in melting and boiling points of the given compounds have to be accounted.

Concept Introduction:

Intermolecular forces in a molecule are the important factor which is responsible for the physical changes of the molecule.

The intermolecular forces in gases are negligibly small and the forces are comparatively stronger in liquids and the intermolecular forces are highest in solids.

The stronger the intermolecular forces, closer will be the molecules in contact and thus higher will be the boiling point and melting point of the substance.

Hydrogen bonding is a special type of dipole-dipole interaction in a polar bond which has hydrogen atom and a highly electronegative atom such as Nitrogen, Oxygen and Fluorine. Hydrogen bonding can also be defined as coulombic attraction between the hydrogen atom and an electronegative atom.

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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…

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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.