The molar mass of oxygen atoms has to be chosen from the given options, given that the atomic mass of oxygen atom is 16.00 a m u . Concept Introduction: Sum of atomic masses of all the atoms that are represented in chemical formula of substance is known as formula mass. If atomic mass and chemical formula of a compound is known, then formula mass can be calculated. Molar mass is the mass that is expressed in grams of a substance that is equal to the formula mass of the substance numerically. Molar mass of an element is unique for the particular element. Two natural elements can never have the same molar mass. Mole is defined as the amount of substance that has as many elementary particles as there are atoms in exactly 12 grams of C 6 12 . The relationship between the atomic mass unit and grams unit is, 6.02 x 1023 a m u = 1.00 g o r 1 a m u = 1.66 x 10 − 24 g

Question

Want to see the full answer?

Answer 1

Question

Chapter 6.3, Problem 2QQ

Interpretation Introduction

Interpretation:

The molar mass of oxygen atoms has to be chosen from the given options, given that the atomic mass of oxygen atom is 16.00 amu.

Concept Introduction:

Sum of atomic masses of all the atoms that are represented in chemical formula of substance is known as formula mass. If atomic mass and chemical formula of a compound is known, then formula mass can be calculated.

Molar mass is the mass that is expressed in grams of a substance that is equal to the formula mass of the substance numerically. Molar mass of an element is unique for the particular element. Two natural elements can never have the same molar mass.

Mole is defined as the amount of substance that has as many elementary particles as there are atoms in exactly 12 grams of C612.

The relationship between the atomic mass unit and grams unit is,

6.02 x 1023 amu = 1.00 g or 1 amu = 1.66 x 10−24 g

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

. Consider a base substitution mutation that occurred in a DNA sequence that resulted in a change in the encoded protein from the amino acid glutamic acid to aspartic acid. Normally the glutamic acid amino acid is located on the outside of the soluble protein but not near an active site. O-H¨ A. What type of mutation occurred? O-H B. What 2 types of chemical bonds are found in the R-groups of each amino acid? The R groups are shaded. CH2 CH2 CH2 H2N-C-COOH H2N-C-COOH 1 H Glutamic acid H Aspartic acid C. What 2 types of bonds could each R-group of each of these amino acids form with other molecules? D. Consider the chemical properties of the two amino acids and the location of the amino acid in the protein. Explain what effect this mutation will have on this protein's function and why.

engineered constructs that consist of hollow fibers are acting as synthetic capillaries, around which cells have been loaded. The cellular space around a single fiber can be modeled as if it were a Krogh tissue cylinder. Each fiber has an outside “capillary” radius of 100 µm and the “tissue” radius can be taken as 200 µm. The following values apply to the device:R0 = 20 µM/secaO2 = 1.35 µM/mmHgDO2,T = 1.67 x 10-5 cm2/secPO2,m = 4 x 10-3 cm/secInstead of blood inside the fibers, the oxygen transport and tissue consumption are being investigated by usingan aqueous solution saturated with pure oxygen. As a result, there is no mass transfer resistance in the synthetic“capillary”, only that due to the membrane itself. Rather than accounting for pO2 variations along the length ofthe fiber, use an average value in the “capillary” of 130 mmHg.Is the tissue fully oxygenated?

Molecular Biology Please help with question. thank you You are studying the expression of the lac operon. You have isolated mutants as described below. In the presence of glucose, explain/describe what would happen, for each mutant, to the expression of the lac operon when you add lactose AND what would happen when the bacteria has used up all of the lactose (if the mutant is able to use lactose).5. Mutations in the lac operator that strengthen the binding of the lac repressor 200 fold 6. Mutations in the promoter that prevent binding of RNA polymerase 7. Mutations in CRP/CAP protein that prevent binding of cAMP8. Mutations in sigma factor that prevent binding of sigma to core RNA polymerase