(a) Interpretation: The number of molecules present in 3.45 g of C 6 H 12 O 6 should be calculated. Concept Introduction: According to Avogadro’s number , one mole of a compound contains 6.023 × 10 23 molecules. Number of moles of a compound is calculated by dividing the given mass of the compound by its molar mass. Moles = M a s s M o l a r M a s s .
(a) Interpretation: The number of molecules present in 3.45 g of C 6 H 12 O 6 should be calculated. Concept Introduction: According to Avogadro’s number , one mole of a compound contains 6.023 × 10 23 molecules. Number of moles of a compound is calculated by dividing the given mass of the compound by its molar mass. Moles = M a s s M o l a r M a s s .
Solution Summary: The author explains how the number of molecules in a compound is calculated by dividing the given mass of the compound by its molar mass.
Definition Definition Number of atoms/molecules present in one mole of any substance. Avogadro's number is a constant. Its value is 6.02214076 × 10 23 per mole.
Chapter 8, Problem 116AP
Interpretation Introduction
(a)
Interpretation:
The number of molecules present in 3.45 g of C6H12O6 should be calculated.
Concept Introduction:
According to Avogadro’s number, one mole of a compound contains 6.023×1023 molecules.
Number of moles of a compound is calculated by dividing the given mass of the compound by its molar mass.
Moles=MassMolarMass.
Interpretation Introduction
(b)
Interpretation:
The number of molecules present in 3.45 moles of C6H12O6 should be calculated.
Concept Introduction:
According to Avogadro’s number, one mole of a compound contains 6.023×1023 molecules.
Interpretation Introduction
(c)
Interpretation:
The number of molecules present in 25 g of ICl5 should be calculated.
Concept Introduction:
According to Avogadro’s number, one mole of a compound contains 6.023×1023 molecules.
Number of moles of a compound is calculated by dividing the given mass of the compound by its molar mass.
Moles=MassMolarMass.
Interpretation Introduction
(d)
Interpretation:
The number of molecules present in 1g of B2H6 should be calculated.
Concept Introduction:
According to Avogadro’s number, one mole of a compound contains 6.023×1023 molecules.
Number of moles of a compound is calculated by dividing the given mass of the compound by its molar mass.
Moles=MassMolarMass.
Interpretation Introduction
(e)
Interpretation:
The number of molecules present in 1.05 mmol of Al(NO3)3 should be calculated.
Concept Introduction:
According to Avogadro’s number, one mole of a compound contains 6.023×1023 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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Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY