CHEMISTRY IN FOCUS W/ OWL (LL)>IP<
6th Edition
ISBN: 9781337306317
Author: Tro
Publisher: CENGAGE L
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Chapter 6, Problem 24E
Interpretation Introduction
Interpretation:
The chlorofluorocarbons, their uses, and their threat to the environment are to be described.
Concept Introduction:
Compounds that consist of a carbon atom are considered organic compounds and the study of these carbon consisting compounds is known as
An organic compound is considered a hydrocarbon if it consists of only carbon and hydrogen atoms. The atoms in a hydrocarbon compound are linked by a single, double, or triple bond.
A group of atoms accountable for the specific reactions in order to form a particular compound is known as
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Solve for x, where M is molar and s is seconds.
x = (9.0 × 10³ M−². s¯¹) (0.26 M)³
Enter the answer. Include units. Use the exponent key above the answer box to indicate any exponent on your units.
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Learning Goal:
This question reviews the format for writing an element's written symbol. Recall that written symbols have a particular format. Written symbols use a form like this:
35 Cl
17
In this form the mass number, 35, is a stacked superscript. The atomic number, 17, is a stacked subscript. "CI" is the chemical symbol for the element chlorine. A general way to show this form is:
It is also correct to write symbols by leaving off the atomic number, as in the following form:
atomic number
mass number Symbol
35 Cl or
mass number Symbol
This is because if you write the element symbol, such as Cl, you know the atomic number is 17 from that symbol. Remember that the atomic number, or number of protons in the nucleus, is what defines the element. Thus, if 17 protons
are in the nucleus, the element can only be chlorine. Sometimes you will only see 35 C1, where the atomic number is not written.
Watch this video to review the format for written symbols.
In the following table each column…
need help please and thanks dont understand only need help with C-F
Learning Goal:
As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7).
Part A - Difference in binding free eenergies
Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol.
The margin of error is 2%.
Part B - Compare difference in free energy to the thermal…
Chapter 6 Solutions
CHEMISTRY IN FOCUS W/ OWL (LL)>IP<
Ch. 6 - Drawing Structural and Condensed Structural...Ch. 6 - Prob. 6.2YTCh. 6 - Drawing Structural Formulas for Isomers Draw...Ch. 6 - Prob. 6.4YTCh. 6 - What are the names of the straight-chain alkanes...Ch. 6 - Which structure corresponds to CH2=CHCH3?Ch. 6 - For the following molecule, why is the name...Ch. 6 - What property is characteristic of chlorinated...Ch. 6 - Prob. 5SCCh. 6 - Prob. 6SC
Ch. 6 - Prob. 7SCCh. 6 - Prob. 1ECh. 6 - Prob. 2ECh. 6 - Prob. 3ECh. 6 - Prob. 4ECh. 6 - What is vitalism? Why did vitalism become a...Ch. 6 - Prob. 6ECh. 6 - Prob. 7ECh. 6 - Prob. 8ECh. 6 - Prob. 9ECh. 6 - List four common fuels used by our society, and...Ch. 6 - Prob. 11ECh. 6 - Why are alkenes and alkynes called unsaturated...Ch. 6 - Prob. 13ECh. 6 - Prob. 14ECh. 6 - Prob. 15ECh. 6 - Prob. 16ECh. 6 - Prob. 17ECh. 6 - Prob. 18ECh. 6 - Prob. 19ECh. 6 - Prob. 20ECh. 6 - Prob. 21ECh. 6 - Prob. 22ECh. 6 - Prob. 23ECh. 6 - Prob. 24ECh. 6 - Prob. 25ECh. 6 - Prob. 26ECh. 6 - Prob. 27ECh. 6 - Prob. 28ECh. 6 - Prob. 29ECh. 6 - Prob. 30ECh. 6 - Prob. 31ECh. 6 - Prob. 32ECh. 6 - Prob. 33ECh. 6 - Prob. 34ECh. 6 - Prob. 35ECh. 6 - Prob. 36ECh. 6 - Prob. 37ECh. 6 - Prob. 38ECh. 6 - Prob. 39ECh. 6 - Prob. 40ECh. 6 - Prob. 41ECh. 6 - Prob. 42ECh. 6 - Naming Hydrocarbons Name each alkane:Ch. 6 - Name each alkane:Ch. 6 - Name each alkyne:Ch. 6 - Name each alkyne: a.CH3CHCHCH2CH2CH3Ch. 6 - Name each alkyne:Ch. 6 - Prob. 48ECh. 6 - Prob. 49ECh. 6 - Prob. 50ECh. 6 - Drawing Hydrocarbon Structures from Names Draw the...Ch. 6 - Draw the condensed structural formula for each...Ch. 6 - Prob. 53ECh. 6 - Prob. 54ECh. 6 - Prob. 55ECh. 6 - Prob. 56ECh. 6 - Functionalized Hydrocarbons Identify each compound...Ch. 6 - Identify each compound according to its functional...Ch. 6 - Identify each compound according to its functional...Ch. 6 - Identify each compound according to its functional...Ch. 6 - Propane, CH3CH2CH3, is a gas at room temperature,...Ch. 6 - Prob. 62ECh. 6 - What was the impact of vitalisms downfall on...Ch. 6 - Why do you think our society has mixed feelings...Ch. 6 - Prob. 65ECh. 6 - Prob. 66ECh. 6 - Prob. 67ECh. 6 - Any one molecule can be represented many ways. For...Ch. 6 - Explain why the formula CH3CH2CH3 cannot mean:...Ch. 6 - Prob. 70ECh. 6 - Prob. 71ECh. 6 - Prob. 72ECh. 6 - Prob. 73E
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- need help please and thanks dont understand only need help with C-F Learning Goal: As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT. The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7). Part A - Difference in binding free eenergies Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol. The margin of error is 2%. Part B - Compare difference in free energy to the thermal…arrow_forwardPlease correct answer and don't used hand raitingarrow_forwardneed help please and thanks dont understand a-b Learning Goal: As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT. The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7). Part A - Difference in binding free eenergies Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol. The margin of error is 2%. Part B - Compare difference in free energy to the thermal energy Divide the…arrow_forward
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