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Concept explainers
(a)
Interpretation:
A balanced chemical equation for the explosive decomposition of liquid nitroglycerine to form carbon dioxide, water vapor, nitrogen, and oxygen is to be written.
Concept introduction:
In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.
Following are the steps to write a balanced chemical equation.
Step 1: Translate the chemical statement into a skeleton equation. The chemical substances that undergo a change are termed as reactants and the chemical substances that are produced during the chemical change are termed as products. The reactants are specified on the left side of the yield arrow while the products are specified on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.
Step 2: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.
Step 3: Balance the remaining atoms by placing the
Step 4: In a balanced
Step 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.
Step 6: Specify the
(b)
Interpretation:
A balanced chemical equation for the reaction between solid potassium superoxide and carbon dioxide to form oxygen gas and solid potassium carbonate is to be written.
Concept introduction:
In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.
Following are the steps to write a balanced chemical equation.
Step 1: Translate the chemical statement into a skeleton equation. The chemical substances that undergo a change are termed as reactants and the chemical substances that are produced during the chemical change are termed as products. The reactants are specified on the left side of the yield arrow while the products are specified on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.
Step 2: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.
Step 3: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it.
Step 4: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.
Step 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.
Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation. The table for the abbreviations used for each state is as follows:
(c)
Interpretation:
A balanced chemical equation for the reaction in a blast furnace between solid iron(III) oxide and carbon monoxide to form solid iron metal and carbon dioxide gas is to be written.
Concept introduction:
In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.
Following are the steps to write a balanced chemical equation.
Step 1: Translate the chemical statement into a skeleton equation. The chemical substances that undergo a change are termed as reactants and the chemical substances that are produced during the chemical change are termed as products. The reactants are specified on the left side of the yield arrow while the products are specified on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.
Step 2: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.
Step 3: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it.
Step 4: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.
Step 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.
Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation. The table for the abbreviations used for each state is as follows:
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Chapter 3 Solutions
Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
- This deals with synthetic organic chemistry. Please fill in the blanks appropriately.arrow_forwardUse the References to access important values if needed for this question. What is the IUPAC name of each of the the following? 0 CH3CHCNH₂ CH3 CH3CHCNHCH2CH3 CH3arrow_forwardYou have now performed a liquid-liquid extraction protocol in Experiment 4. In doing so, you manipulated and exploited the acid-base chemistry of one or more of the compounds in your mixture to facilitate their separation into different phases. The key to understanding how liquid- liquid extractions work is by knowing which layer a compound is in, and in what protonation state. The following liquid-liquid extraction is different from the one you performed in Experiment 4, but it uses the same type of logic. Your task is to show how to separate apart Compound A and Compound B. . Complete the following flowchart of a liquid-liquid extraction. Handwritten work is encouraged. • Draw by hand (neatly) only the appropriate organic compound(s) in the boxes. . Specify the reagent(s)/chemicals (name is fine) and concentration as required in Boxes 4 and 5. • Box 7a requires the solvent (name is fine). • Box 7b requires one inorganic compound. • You can neatly complete this assignment by hand and…arrow_forward
- b) Elucidate compound D w) mt at 170 nd shows c-1 stretch at 550cm;' The compound has the ff electronic transitions: 0%o* and no a* 1H NMR Spectrum (CDCl3, 400 MHz) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 13C{H} NMR Spectrum (CDCl3, 100 MHz) Solvent 80 70 60 50 40 30 20 10 0 ppm ppm ¹H-13C me-HSQC Spectrum ppm (CDCl3, 400 MHz) 5 ¹H-¹H COSY Spectrum (CDCl3, 400 MHz) 0.5 10 3.5 3.0 2.5 2.0 1.5 1.0 10 15 20 20 25 30 30 -35 -1.0 1.5 -2.0 -2.5 3.0 -3.5 0.5 ppm 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppmarrow_forwardShow work with explanation. don't give Ai generated solutionarrow_forwardRedraw the flowchartarrow_forward
- redraw the flowchart with boxes and molecules written in themarrow_forwardPart I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward• Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forward
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