Calculate the number of molecules present in each of the following samples. l type="a"> 4.75 mmol of phosphine, PH 3 i>4.75 g of phosphine, PH 3 i> 1.25 × 10 − 2 g of lead(II) acetate, Pb ( CH 3 CO 2 ) 2 i>1.25 X 10 2 moles of lead(II) acetate, Pb ( CH 3 CO 2 ) 2 i>a sample of benzene, C 6 H 6 , which contains a total of 5.40 moles of carbon
Calculate the number of molecules present in each of the following samples. l type="a"> 4.75 mmol of phosphine, PH 3 i>4.75 g of phosphine, PH 3 i> 1.25 × 10 − 2 g of lead(II) acetate, Pb ( CH 3 CO 2 ) 2 i>1.25 X 10 2 moles of lead(II) acetate, Pb ( CH 3 CO 2 ) 2 i>a sample of benzene, C 6 H 6 , which contains a total of 5.40 moles of carbon
Therefore, number of molecules in lead (II) acetate is 7.53×1021 molecules.
Expert Solution
Interpretation Introduction
(e)
Interpretation:
The number of molecules of substance should be calculated.
Concept Introduction:
Number of moles is related to mass and molar mass as follows:
n=mM
Here, m is mass and M is molar mass.
According to Avogadro’s law, 1 mole of a substance contains 6.023×1023 atoms. This is known as Avogadro’s number and denoted by symbol NA.
Thus, number of molecules can be calculated from number of moles using the following conversion factor:
(6.023×1023 molecules1 mol).
Answer to Problem 39QAP
5.42×1023 molecules.
Explanation of Solution
Total number of moles of carbon is 5.40 mol.
From the formula of benzene C6H6, 1 mol of benzene contains 6 mol of carbon thus, 1 mol of carbon will be present in 1/6 moles of benzene and 5.40 mol will be present in 16×5.40 mol=0.9 mol
Since, 1 mole of benzene contains 6.023×1023 molecules thus, 0.9 mol of benzene will have following molecules:
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…
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…
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