(a)
Interpretation: To name and determine no. of neutrons in the isotope
Concept introduction:
Isotopes are two or more different chemical forms of the same element that have the same number of protons but varying numbers of neutrons in their nucleus.
(a)
Answer to Problem 99A
Name of the isotope is silver
No. of neutrons is 62
Explanation of Solution
Given isotope:
Name of the element is silver.
The
The mass number is 109.
Expression for mass number is,
So, calculate no. of neutrons as shown,
So, no. of neutrons is 62.
(b)
Interpretation: To name and determine no. of electrons in the isotope
Concept introduction:
Isotopes are any two or more different chemical forms of the same element that have the same number of protons but varying numbers of neutrons in their nucleus.
(b)
Answer to Problem 99A
Name of the isotope is tin.
No. of protons is 50.
Explanation of Solution
Given isotope:
The atomic number is 50. The atomic number is no. of protons.
So, no. of protons is 50.
(c)
Interpretation: To determine the number of isotopes in an isotope.
Concept introduction:
Isotopes are any two or more different chemical forms of the same element that have the same number of protons but varying numbers of neutrons in their nucleus.
(c)
Answer to Problem 99A
Name of the isotope molybdenum.
No. of electrons is 42.
Explanation of Solution
Given isotope:
The atomic number is 42. The atomic number is no. of protons.
In any element no. of protons and neutrons are the same.
So, no. of electrons is 42.
(d)
Interpretation: To determine the number of isotopes in an isotope.
Concept introduction:
Isotopes are any two or more different chemical forms of the same element that have the same number of protons but varying numbers of neutrons in their nucleus.
(d)
Answer to Problem 99A
Name of the isotope scandium.
No. of electrons is 21.
Explanation of Solution
Given isotope:
The atomic number is 21. The atomic number is no. of protons.
In any element no. of protons and neutrons are the same.
So, no. of electrons is 21.
Chapter 6 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- The acid-base indicator HX undergoes the following reaction in a dilute aqueous solution: HX (color 1) H+ + X- (color 2). The following absorbance data were obtained for a 0.00035 M solution of HX in 0.1 M NaOH and 0.1 M HCI. Measurements were made at wavelengths of 450 nm and 620 nm using a 1.0 cm glass cuvette. 450 620 A(460 nm) A(630 nm) 0.1 M NaOH 0.1 M HCI 0.065 0.435 0.895 0.150 In the 0.1M NaOH solution, the indicator will be almost 100% in the X- form, while in 0.1M HCI, the indicator will be nearly 100% protonated (HX). Calculate the acid dissociation constant for the indicator if a pH=5 buffer solution containing a very small amount of indicator exhibits an absorbance of 0.567 at 450 nm and 0.395 at 620 nm (measured in a 1 cm glass cuvette).arrow_forwardShow work...give the name of the given compound. Don't give Ai generated solutionarrow_forwardShow work with explanation needed. don't give Ai generated solutionarrow_forward
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