The number of moles of compound is given. By using the number of moles, the mass of nitrogen in each compound given in exercise 51 is to be determined. Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons. Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound. The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it. The amount of substance containing 12 g of pure carbon is called a mole. One mole of substance always contains 6 .022 × 10 23 atoms. Hence, ( 6 .022 × 10 23 atoms ) ( 12 u 1 atom ) = 12 g ⇒ 1 u = 1 6 .022 × 10 23 g To determine : The mass of nitrogen ( N ) in 5 .00 moles of NH 3 .
The number of moles of compound is given. By using the number of moles, the mass of nitrogen in each compound given in exercise 51 is to be determined. Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons. Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound. The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it. The amount of substance containing 12 g of pure carbon is called a mole. One mole of substance always contains 6 .022 × 10 23 atoms. Hence, ( 6 .022 × 10 23 atoms ) ( 12 u 1 atom ) = 12 g ⇒ 1 u = 1 6 .022 × 10 23 g To determine : The mass of nitrogen ( N ) in 5 .00 moles of NH 3 .
Solution Summary: The author explains that the number of moles of a compound is given and the mass of nitrogen in each compound given in exercise 51 is determined.
Interpretation: The number of moles of compound is given. By using the number of moles, the mass of nitrogen in each compound given in exercise 51 is to be determined.
Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.
Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.
The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.
The amount of substance containing
12g of pure carbon is called a mole. One mole of substance always contains
6.022×1023 atoms.
Hence,
(6.022×1023atoms)(12u1atom)=12g⇒1u=16.022×1023g
To determine: The mass of nitrogen
(N) in
5.00 moles of
NH3.
(b)
Interpretation Introduction
Interpretation: The number of moles of compound is given. By using the number of moles, the mass of nitrogen in each compound given in exercise 51 is to be determined.
Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.
Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.
The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.
The amount of substance containing
12g of pure carbon is called a mole. One mole of substance always contains
6.022×1023 atoms.
Hence,
(6.022×1023atoms)(12u1atom)=12g⇒1u=16.022×1023g
To determine: The mass of nitrogen
(N) in
5.00 moles of
N2H4.
(c)
Interpretation Introduction
Interpretation: The number of moles of compound is given. By using the number of moles, the mass of nitrogen in each compound given in exercise 51 is to be determined.
Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.
Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.
The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.
The amount of substance containing
12g of pure carbon is called a mole. One mole of substance always contains
6.022×1023 atoms.
Hence,
(6.022×1023atoms)(12u1atom)=12g⇒1u=16.022×1023g
To determine: The mass of nitrogen
(N) in
5.00 moles of
(NH4)2Cr2O7.
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…
Please correct answer and don't used hand raiting
need 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…
Chapter 5 Solutions
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