
Concept explainers
Interpretation:
The given chemical equations are need to be balanced.
Concept Introduction:
Chemical equations are symbolic representations of
The reactants are written on the left side whereas the products are written on the right side of an arrow.
Chemical equations are denoted using chemical formulae of the elements and compounds involved.
The
Atomicity is defined as the number of atoms in a molecule of an element.
Chemical formulae of compounds are written on the basis of their molecularity.
Molecularity is the number of molecules that come to react in an elementary (single-step) reaction.
A balanced chemical equation shows the same number of atoms of each element on both sides of the arrow.
Chemical equations are balanced on the basis of the Law of Conservation of Mass.

Answer to Problem 25QP
Solution:
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
Explanation of Solution
a)
There are two atoms of nitrogen on either side of the equation. However, there are five oxygen atoms on the left side and six oxygen atoms on the right side of the equation. The entire equation is balanced by adding the coefficient
b)
There is one atom of potassium on either side of the equation. Also, there is one atom of nitrogen on both the sides. However, on the left there are three oxygen atoms while on the right, there are four oxygen atoms. The entire equation is balanced by adding the coefficient
c)
There are two atoms of nitrogen on either side of the equation. However, there is an imbalance in the number of hydrogen and oxygen atoms. In case of hydrogen, there are four atoms on the left side and two on the right side. In case of oxygen, there are three atoms on the left side and two on the right side. The entire equation is balanced by adding the coefficient
d)
There are two atoms of nitrogen on either side of the equation. However, there are four atoms of hydrogen on the left side and two atoms on the right side. Also, there are two oxygen atoms on the left side and one on the right side of the equation. The entire equation is balanced by adding the coefficient
e)
There is one atom of sodium on the left side while two atoms on the right side of the equation. For hydrogen, there is one atom on the left but two atoms on the right. For carbon also, there is one atom in the left but two atoms on the right. For oxygen, there are three atoms on the left but six atoms on the right of the equation. The entire equation is balanced by adding the coefficient
f)
There are four atoms of phosphorus on the left side but only one on the right side of the equation. In case of hydrogen, there are two atoms on the left side and three on the right. In case of oxygen, there are eleven atoms on the left but four on the right. The entire equation is balanced by adding the coefficient
g)
There is only one atom of calcium on either side of the equation. It is the same case with the elements carbon and oxygen, where it is one carbon atom and three oxygen atoms on the either side of the equation. However, there are discrepancies in the number of hydrogen and chlorine atoms. In case of chlorine, there is one on the left side and two on the right side. In case of hydrogen, there is one atom on the left side but two on the right side. The entire equation is balanced by adding the coefficient
h)
Apart from hydrogen, the number of all other atoms is different on the either side of the equation. In case of hydrogen, there are two atoms on both sides of the equation. On the other hand, for aluminium, there is one atom on the left and two atoms on the right. For sulphur, there is one atom on the left but three on the right. For oxygen, there are four atoms on the left but twelve on the right. The entire equation is balanced by adding the coefficient
i)
There is one atom of carbon on both sides of the equation. For potassium, there is one atom on the left but two on the right side of the equation. For hydrogen, there is one atom on the left side and two on the right side of the equation. For oxygen, there is one atom on the left side but four on the right side. The entire equation is balanced by adding the coefficient
j)
There is one atom of carbon on both sides of the equation. For hydrogen, there are four atoms on the left side and two on the right side of the equation. For oxygen, there are two atoms on the left side but three on the right side. The entire equation is balanced by adding the coefficient
k)
There is one atom of carbon on both sides of the equation. For hydrogen, there are two atoms on the left side but six on the right side of the equation. For oxygen, there is one atom on the left side but two on the right side. For beryllium, there is one atom on the left side but one on the right side. The entire equation is balanced by adding the coefficient
l)
There is one atom of copper on both sides of the equation. For oxygen, there are three atoms on the left side but eight on the right side of the equation. For hydrogen, there is one atom on the left side but two on the right side. For nitrogen, there is one atom on the left side but three on the right side. The entire equation is balanced by adding the coefficient
m)
There is one atom of sulphur on both sides of the equation. For oxygen, there are three atoms on the left side but seven on the right side of the equation. For hydrogen, there is one atom on the left side but four on the right side. For nitrogen, there is one atom on the left side and one on the right side. The entire equation is balanced by adding the coefficient
n)
There is one atom of copper on both sides of the equation. It is the same for oxygen. For hydrogen, there are three atoms on the left side but two on the right side. For nitrogen, there is one atom on the left side but two on the right side. The entire equation is balanced by adding the coefficient
Want to see more full solutions like this?
Chapter 3 Solutions
CHEMISTRY >CUSTOM<
- Given a complex reaction with rate equation v = k1[A] + k2[A]2, what is the overall reaction order?arrow_forwardPlease draw the structure in the box that is consistent with all the spectral data and alphabetically label all of the equivalent protons in the structure (Ha, Hb, Hc....) in order to assign all the proton NMR peaks. The integrations are computer generated and approximate the number of equivalent protons. Molecular formula: C13H1802 14 13 12 11 10 11 (ppm) Structure with assigned H peaks 2.08 3.13arrow_forwardCHEMICAL KINETICS. One of the approximation methods for solving the rate equation is the steady-state approximation method. Explain what it consists of.arrow_forward
- CHEMICAL KINETICS. One of the approximation methods for solving the rate equation is the limiting or determining step approximation method. Explain what it consists of.arrow_forwardCHEMICAL KINETICS. Indicate the approximation methods for solving the rate equation.arrow_forwardTRANSMITTANCE เบบ Please identify the one structure below that is consistent with the 'H NMR and IR spectra shown and draw its complete structure in the box below with the protons alphabetically labeled as shown in the NMR spectrum and label the IR bands, including sp³C-H and sp2C-H stretch, indicated by the arrows. D 4000 OH LOH H₂C CH3 OH H₂C OCH3 CH3 OH 3000 2000 1500 HAVENUMBERI-11 1000 LOCH3 Draw your structure below and label its equivalent protons according to the peak labeling that is used in the NMR spectrum in order to assign the peaks. Integrals indicate number of equivalent protons. Splitting patterns are: s=singlet, d=doublet, m-multiplet 8 3Hb s m 1Hd s 3Hf m 2Hcd 2Had 1He 鄙视 m 7 7 6 5 4 3 22 500 T 1 0arrow_forward
- Relative Transmittance 0.995 0.99 0.985 0.98 Please draw the structure that is consistent with all the spectral data below in the box and alphabetically label the equivalent protons in the structure (Ha, Hb, Hc ....) in order to assign all the proton NMR peaks. Label the absorption bands in the IR spectrum indicated by the arrows. INFRARED SPECTRUM 1 0.975 3000 2000 Wavenumber (cm-1) 1000 Structure with assigned H peaks 1 3 180 160 140 120 100 f1 (ppm) 80 60 40 20 0 C-13 NMR note that there are 4 peaks between 120-140ppm Integral values equal the number of equivalent protons 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 fl (ppm)arrow_forwardCalculate the pH of 0.0025 M phenol.arrow_forwardIn the following reaction, the OH- acts as which of these? NO2-(aq) + H2O(l) ⇌ OH-(aq) + HNO2(aq)arrow_forward
- Using spectra attached, can the unknown be predicted? Draw the predicition. Please explain and provide steps. Molecular focrmula:C16H13ClOarrow_forwardCalculate the percent ionization for 0.0025 M phenol. Use the assumption to find [H3O+] first. K = 1.0 x 10-10arrow_forwardThe Ka for sodium dihydrogen phosphate is 6.32 x 10-8. Find the pH of a buffer made from 0.15 M H2PO4- and 0.25 M HPO42- .arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
- Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning





