EBK CHEMICAL PRINCIPLES IN THE LABORATO
11th Edition
ISBN: 9780100546066
Author: SLOWINSKI
Publisher: YUZU
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Chapter 17, Problem 3ASA
A given substance is a white, granular solid at
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A buffer is prepared by adding 0.50 mol of acetic acid (HC2H3O2) and 0.75 mol of sodium acetate
(NaC2H3O2) to enough water to form 2.00L solution. (pKa for acetic acid is 4.74) Calculate the pH of the
buffer.
Modify the given carbon skeleton to draw the major product of the following reaction. If a racemic mixture of enantiomers is
expected, draw both enantiomers. Note: you can select a structure and use Copy and Paste to save drawing time.
HBr
کی
CH3
کی
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Sort the following into the classification for a reaction that is NOT at equilibrium versus a reaction system that has reached equilibrium.
Drag the appropriate items to their respective bins.
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The forward and reverse reactions
proceed at the same rate.
Chemical equilibrium is a dynamic
state.
The ratio of products to reactants is
not stable.
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The state of chemical equilibrium will
remain the same unless reactants or
products escape or are introduced into
the system. This will disturb the
equilibrium.
The concentration of products is
increasing, and the concentration of
reactants is decreasing.
The ratio of products to reactants
does not change.
The rate at which products form from
reactants is equal to the rate at which
reactants form from products.
The concentrations of reactants and
products are stable and cease to
change.
The reaction has reached equilibrium.
The rate of the forward reaction is
greater than the rate of the reverse
reaction.
The…
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EBK CHEMICAL PRINCIPLES IN THE LABORATO
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- Place the following characteristics into the box for the correct ion. Note that some of the characteristics will not be placed in either bin. Use your periodic table for assistance. Link to Periodic Table Drag the characteristics to their respective bins. ▸ View Available Hint(s) This anion could form a neutral compound by forming an ionic bond with one Ca²+. Reset Help This ion forms ionic bonds with nonmetals. This ion has a 1- charge. This is a polyatomic ion. The neutral atom from which this ion is formed is a metal. The atom from which this ion is formed gains an electron to become an ion. The atom from which this ion is formed loses an electron to become an ion. This ion has a total of 18 electrons. This ion has a total of 36 electrons. This ion has covalent bonds and a net 2- charge. This ion has a 1+ charge. Potassium ion Bromide ion Sulfate ionarrow_forwardU Consider the following graph containing line plots for the moles of Product 1 versus time (minutes) and the moles of Product 2 versus time in minutes. Choose all of the key terms/phrases that describe the plots on this graph. Check all that apply. ▸ View Available Hint(s) Slope is zero. More of Product 1 is obtained in 12 minutes. Slope has units of moles per minute. plot of minutes versus moles positive relationship between moles and minutes negative relationship between moles and minutes Slope has units of minutes per moles. More of Product 2 is obtained in 12 minutes. can be described using equation y = mx + b plot of moles versus minutes y-intercept is at (12,10). y-intercept is at the origin. Product Amount (moles) Product 1 B (12,10) Product 2 E 1 Time (minutes) A (12,5)arrow_forwardSolve 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. ▸ View Available Hint(s) ΜΑ 0 ? Units Valuearrow_forward
- 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…arrow_forwardneed 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_forwardneed 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_forward
- Please 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_forwardPlease correct answer and don't used hand raitingarrow_forward
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