The answer to the riddle Maggie is solving is 6
To determine the answer to the riddle Maggie is solving,
We will subtract the number of valence electrons in halogens by the number of valence electrons in alkali metals.
First, let us define what is meant by valence electrons
Valence electrons are those electrons that reside in the outermost shell surrounding an atomic nucleus.
Now, we will the determine the respective number of valence electrons in halogens as well as the number of valence electrons in alkali metals.
For Halogens
Halogens are the group 17 (group 7A) elements in the periodic table. Examples are Fluorine, Chlorine, Bromine etc.
Halogens have 7 electrons in their outermost shell
For Alkali metals
Alkali metals are the group 1 metals. Examples are Lithium, Sodium, Potassium etc.
Alkali metals have 1 electron in their outermost shell
Now, for the answer to the riddle,
Answer to the riddle = Number of valence electrons in halogens - Number of valence electrons in Alkali metals
\
Answer to the riddle = 7 - 1
∴ Answer to the riddle = 6
Hence, the answer to the riddle Maggie is solving is 6
Learn more here: brainly.com/question/9392364
It dissolves.
It increases the pH of the solution.
It releases hydroxide ions.
Answer: Option (a) is the correct answer.
Explanation:
An acid-base indicator is an indicator which changes color according to the pH of solution in which it is placed.
For example, phenolphthalein is an acid-base indicator.
Thus, we can conclude that out of the given options, an acid–base indicator react by changing color when placed in an acidic solution, is the correct option.
B. an inference.
C. a theory.
D. a controlled experiment.
Answer:
Detail is given below.
Explanation:
Given data;
Abundance of Berkmarium-95 = 70%
Abundance of Berkmarium-97 = 28%
Abundance of Berkmarium-94 = 2%
Average atomic mass closer to which isotope = ?
Solution:
1st of all we will calculate the average atomic mass of Berkmarium.
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (70×95)+(28×97)+(2×94) /100
Average atomic mass = 6650 + 2716+ 188 / 100
Average atomic mass= 9554 / 100
Average atomic mass = 95.54 amu
The average atomic mass is closer to the isotope Berkmarium-95 because it is present in abundance as compared to the other two isotope. So this isotope constitute most of the part of Berkmarium.