Interpretation : The number of protons for the given element is to be calculated.
Concept Introduction : The positively charged subatomic particles known as protons were first found in the nucleus by Eugen Goldstein.
Answer to Problem 17SP
The number of protons in the element of
Explanation of Solution
Given information:
The atomic number of sulphur is 16.
The positively charged subatomic particles known as protons were first found in the nucleus by Eugen Goldstein.
The count of protons in the nucleus of an element's atom is known as the element's atomic number. An element is identified by its atomic number.
As the atomic number of sulphur is 16, the neutral sulphur atom has 16 protons.
Interpretation : The number of electrons for the given element is to be calculated.
Concept Introduction : Electrons are negatively charged subatomic particles that revolve around the nucleus discovered by J. J. Thomson.
Answer to Problem 17SP
The number of electrons in the element of atomic number 16 is 16.
Explanation of Solution
Given information :
The atomic number of sulphur is 16.
Electrons are negatively charged subatomic particles that revolve around the nucleus discovered by J. J. Thomson.
The atomic number of an element is equal to the number of electrons that revolve around the nucleus.
As the atomic number of sulphur is 16, the number of electrons is also 16.
Interpretation : The atomic number for the given element is to be calculated.
Concept Introduction : The count of protons in the nucleus of an element's atom is known as the element's atomic number. An element is identified by its atomic number.
Answer to Problem 17SP
The atomic number for the element is 23.
Explanation of Solution
Given information :
The number of protons is 23.
The count of protons in the nucleus of an element's atom is known as the element's atomic number.
An element is identified by its atomic number.
As the number of protons is 23, the atomic number will also be 23.
Interpretation : The number of electrons for the given element is to be calculated.
Concept Introduction : Electrons are negatively charged subatomic particles that revolve around the nucleus discovered by J. J. Thomson.
Answer to Problem 17SP
The number of electrons for the element is 23.
Explanation of Solution
Given information :
The number of protons is 23.
Electrons are negatively charged subatomic particles that revolve around the nucleus discovered by J. J. Thomson.
The number of protons is equal to the number of electrons for a particular element.
Since the count of protons is 23, the number of electrons is also 23.
Interpretation : The given element is to be calculated.
Concept Introduction : The count of protons in the nucleus of an element's atom is known as the element's atomic number.
Answer to Problem 17SP
The element with atomic number 5 is boron and its symbol is B.
Explanation of Solution
Given information :
The number of electrons is 5.
The count of protons in the nucleus of an element's atom is known as the element's atomic number.
The number of electrons is equal to the element’s atomic number.
The element with atomic number 5 is Boron. The symbol of boron is B.
Interpretation : The atomic number for the given element is to be calculated.
Concept Introduction : The count of protons in the nucleus of an element's atom is known as the element's atomic number.
Answer to Problem 17SP
The atomic number of the element is 5.
Explanation of Solution
Given information :
The number of electrons is 5.
The number of protons in the nucleus of an element's atom is known as the element's atomic number.
The number of electrons is equal to the element’s atomic number.
So, the atomic number of the element is 5.
Interpretation : The number of protons for the given element is to be calculated.
Concept Introduction : The positively charged subatomic particles known as protons were first found in the nucleus by Eugen Goldstein.
Answer to Problem 17SP
The number of protons of the element is 5.
Explanation of Solution
Given information :
The number of electrons is 5.
The positively charged subatomic particles known as protons were first found in the nucleus by Eugen Goldstein.
The number of protons is equal to the number of electrons for a particular element.
Since the count of electrons is 5, the number of protons is also 5.
Chapter 4 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M* = 163 m/z) is formed. In the infrared spectrum, important absorptions appear at 1661, 750 and 690 cm. The 13C NMR and DEPT spectra are provided. Draw the structure of the product as the resonance contributor lacking any formal charges. 13C NMR DEPT 90 200 160 120 80 40 0 200 160 120 80 40 0 DEPT 135 T 200 160 120 80 40 0 Draw the unknown amide. Select Dow Templates More Fragearrow_forwardIdentify the unknown compound from its IR and proton NMR spectra. C4H6O: 'H NMR: 82.43 (1H, t, J = 2 Hz); 8 3.41 (3H, s); 8 4.10 (2H, d, J = 2 Hz) IR: 2125, 3300 cm¹ The C4H6O compound liberates a gas when treated with C2H5 MgBr. Draw the unknown compound. Select Draw с H Templates Morearrow_forwardPlease help with number 6 I got a negative number could that be right?arrow_forward
- 1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.arrow_forwardGive the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.arrow_forwardFive isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane The IR spectra of these five alkenes have the key absorptions (in cm Compound Compound A –912. (§), 994 (5), 1643 (%), 3077 (1) Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption) Compound C Compound D) –714 (5), 1665 (w), 3010 (m) 885 (3), 1650 (m), 3086 (m) 967 (5), no aharption 1600 to 1700, 3040 (m) Compound K Match each compound to the data presented. Compound A Compound B Compound C Compound D Compoundarrow_forward
- 7. The three sets of replicate results below were accumulated for the analysis of the same sample. Pool these data to obtain the most efficient estimate of the mean analyte content and the standard deviation. Lead content/ppm: Set 1 Set 2 Set 3 1. 9.76 9.87 9.85 2. 9.42 9.64 9.91 3. 9.53 9.71 9.42 9.81 9.49arrow_forwardDraw the Zaitsev product famed when 2,3-dimethylpentan-3-of undergoes an El dehydration. CH₂ E1 OH H₁PO₁ Select Draw Templates More QQQ +H₂Oarrow_forwardComplete the clean-pushing mechanism for the given ether synthesia from propanol in concentrated sulfurica140°C by adding any mining aloms, bands, charges, nonbonding electron pairs, and curved arrows. Draw hydrogen bonded to cayan, when applicable. ore 11,0 HPC Step 1: Draw curved arrows Step 2: Complete the intend carved Q2Q 56 QQQ Step 3: Complete the intermediate and add curved Step 4: Modify the structures to draw the QQQ QQQarrow_forward
- 6. In an experiment the following replicate set of volume measurements (cm3) was recorded: (25.35, 25.80, 25.28, 25.50, 25.45, 25.43) A. Calculate the mean of the raw data. B. Using the rejection quotient (Q-test) reject any questionable results. C. Recalculate the mean and compare it with the value obtained in 2(a).arrow_forwardA student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. • If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T G OH де OH This transformation can't be done in one step.arrow_forwardMacmillan Leaming Draw the major organic product of the reaction. 1. CH3CH2MgBr 2. H+ - G Select Draw Templates More H о QQarrow_forward
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY