Concept explainers
Videos
Balance each of the following chemical equations.
l type='a'>
msp;
i>
i>
i>
i>
i>
i>
i>
mg src=Images/HTML_99425-6-41QAP_image001.jpg alt="" align="top"/>
(a)
Interpretation:
The given chemical equation is to be balanced.
Concept Introduction:
In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.
Answer to Problem 41QAP
Explanation of Solution
The provided reaction is:
The most complicated molecule is not clear, so we start arbitrarily. If we place a coefficient of 3 before Fe and 4 before H2 and H2 O, we get a balanced equation.
(b)
Interpretation:
The given chemical equation is to be balanced.
Concept Introduction:
In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.
Answer to Problem 41QAP
Explanation of Solution
The given unbalanced equation is:
We start with placing 2 before KCl and we get the balanced chemical reaction as follows:
(c)
Interpretation:
The given chemical equation is to be balanced.
Concept Introduction:
In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.
Answer to Problem 41QAP
Explanation of Solution
The given unbalanced equation is:
We start by balancing HCl. If we place a coefficient 2 before HCl we get a balanced chemical reaction as follows:
(d)
Interpretation:
The given chemical equation is to be balanced.
Concept Introduction:
In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.
Answer to Problem 41QAP
Explanation of Solution
The provided equation is:
If we place a coefficient of 2 H2 O and HBr, we get a balanced equation as follows:
(e)
Interpretation:
The given chemical equation is to be balanced.
Concept Introduction:
In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.
Answer to Problem 41QAP
Explanation of Solution
The provided equation is:
If we place a coefficient of 3 before Cl2, we get the balanced chemical reaction as follows:
(f)
Interpretation:
The given chemical equation is to be balanced.
Concept Introduction:
In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.
Answer to Problem 41QAP
Explanation of Solution
The provided reaction is:
We can see that all the atoms on both the sides are already balanced. Therefore, we already have a balanced equation.
(g)
Interpretation:
The given chemical equation is to be balanced.
Concept Introduction:
In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.
Answer to Problem 41QAP
Explanation of Solution
The provided reaction is:
The most complicated molecule is
We can balance hydrogen by giving HBr a coefficient of 3. We now have a balanced equation
(h)
Interpretation:
The given chemical equation is to be balanced.
Concept Introduction:
In a balanced chemical reaction, the number of similar type of elements on both sides of the reaction is equal. To balance an unbalanced equation, the coefficients that are present before the compounds are changed but the subscripts in the formulas are not changed.
Answer to Problem 41QAP
Explanation of Solution
The provided reaction is:
Want to see more full solutions like this?
Chapter 6 Solutions
Introductory Chemistry: A Foundation
- In Fourier transformed spectroscopya) use a very sensitive monocromador systemb) the detection time is inferior to conventional spectroscopiac) the signal is detected depending on the frequencyd) occurs simultaneously at all frequency intervalsarrow_forwardIf a radiation intensity l0 = 2.5x1010 fotones s-1cm2 results in a dissolución, an absorption of 0.95 will be recorded. What is the percentage of incident radiation and transmission?a) 88.88% b) 5% c) 11.22% d) 95%arrow_forwardIndicate the spectroscopic transmission that requires greater energy radiation. Justification:a) NMR b) vibration c) electronica d) rotationarrow_forward
- After an induced absorption process of an intensity, there are (without population inversion) transitions between:a) vibrational and rotational levels in the infrared region, we obtainb) vibrational levels exclusively in the infrared regionc) vibrational and rotational levels in the microwave regiond) transitions between nuclear spin levels in the radio frequency regionarrow_forwardIn a spontaneous emission process:a) the ground state population decreasesb) the excited state population decreasesc) the non-radiative component is predominantd) the emitted radiation is coherentarrow_forwardFor a molecule there are 3 energy levels A, B and C, where B is an intermediate energy level between A and C. The A → C transition occurs at 480 nm and the B → C transition occurs at 885 nm. Indicate the wavelength at which the A → B transition will occur.arrow_forward
- For a molecule there are three energy levels: A, B and C. If the transition A → B occurs at 1049 nm and the transition B → C occurs at 885 nm, we can say that the wavelength of the transition A → C will occur at approximately:a) 164 nm b) 1934 nm c) 480 nm d) 967 nmarrow_forward: Naming the Alkanes a) Write the IUPAC nomenclature of the compound below b) Draw 4-isopropyl-2,4,5-trimethylheptane, identify the primary, secondary, tertiary, and quaternary carbons. c) Rank pentane, neopentane and isopentane for boiling point. pentane: H3C-CH2-CH2-CH2-CH3 neopentane: CH3 H3C-C-CH3 isopentane: CH3 CH3 H3C-CH2-CH-CH3arrow_forwardAn essential part of the experimental design process is to select appropriate dependent and independent variables. True Falsearrow_forward
- 10.00 g of Compound X with molecular formula C₂Hg are burned in a constant-pressure calorimeter containing 40.00 kg of water at 25 °C. The temperature of the water is observed to rise by 2.604 °C. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound X at 25 °C. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.arrow_forwardneed help not sure what am doing wrong step by step please answer is 971A During the lecture, we calculated the Debye length at physiological salt concentrations and temperature, i.e. at an ionic strength of 150 mM (i.e. 0.150 mol/l) and a temperature of T=310 K. We predicted that electrostatic interactions are effectively screened beyond distances of 8.1 Å in solutions with a physiological salt concentration. What is the Debye length in a sample of distilled water with an ionic strength of 10.0 µM (i.e. 1.00 * 10-5 mol/l)? Assume room temperature, i.e. T= 298 K, and provide your answer as a numerical expression with 3 significant figures in Å (1 Å = 10-10 m).arrow_forwardInfluence of salt concentrations on electrostatic interactions 2 Answer is 2.17A why not sure step by step please What is the Debye length in a concentrated salt solution with an ionic strength of 2.00 mol/l? Assume room temperature, i.e. T= 298 K, and provide your answer as a numerical expression with 3 significant figures in Å (1 Å = 10-10 m).arrow_forward
General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning
Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning