- The chemical symbol of the isotope boron-11 is ¹¹B.
- The atomic mass of the isotope boron-11 is equal to 11.009306.
- The abundance in nature of the isotope boron-11 is equal to 80.1%.
<h3>What is an isotope?</h3>
An isotope can be defined as the atom of a chemical element that has the same number of protons but different number of neutrons. This ultimately implies that, the isotopes of an element have the same atomic number (number of protons) but different atomic mass (number of neutrons).
In Chemistry, there are two main isotopes of boron and these include the following:
Boron-11 is the most stable isotope of boron and it is characterized by the following:
- The chemical symbol of the isotope boron-11 is ¹¹B.
- The atomic mass of the isotope boron-11 is equal to 11.009306.
- The abundance in nature of the isotope boron-11 is equal to 80.1%.
Read more on Boron-11 here: brainly.com/question/6283234
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Answer:
A. It formed by barium(Ba+2) ion and sulfate ( SO42- )
B. It is formed by calcium ion (Ca+2) and two fluoride ions (2F-)
C. It is formed by magnesium ion (Mg+2) and nitride ion (N3-)
D. It is formed by two potassium ions (2 K+) and oxide ion(O2-)
Answer:
Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.
Explanation:
If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.
The first blank is a compound, second is a mixture
Answer : The time passed in years is 
Explanation :
Half-life of carbon-14 = 5730 years
First we have to calculate the rate constant, we use the formula :
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 
t = time passed by the sample = ?
a = initial amount of the reactant disintegrate = 15.3
a - x = amount left after decay process = 14.8
Now put all the given values in above equation, we get
Therefore, the time passed in years is
