When most radioactive atoms “spontaneously” decay to a more stable form the “additional” energy is converted to radiation with the emission of radioactive particles.
<h3>What is radioactivity?</h3>
Radioactivity is the spontaneous decay of the nucleus of an atom with the emission of radiation and nuclear particles.
Elements that spontaneously decay are called radioactive elements.
When these radioactive elements decay, they form more stable isotopes or elements.
The spontaneous decay of atoms of radioactive elements is in order for the nucleus of the atom to become stable and non-radioactive.
Learn more about radioactivity at:
brainly.com/question/26626062
#SPJ1
Answer:
a. +2
b. +3
c. -1
Explanation:
The typical oxidation states can be determined from the periodic table based on the number of valence electrons an atom has.
a. Calcium belongs to group 2A, meaning it has 2 valence electrons and, therefore, would have an oxidation state of +2 in compounds.
b. Aluminum is in group 3A, meaning it has 3 valence electrons and would have an oxidation state of +3 in compounds when the 3 electrons are lost.
c. Fluorine would become fluorine if it gained 1 additional electron to achieve an octet, so its oxidation state would be -1.
Answer:
KOH and H₂SO₄
Explanation:
Neutralization reaction:
It is the reaction in which acid and base react with each other and produce salt and water.
For example:
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
1. Potassium hydroxide and sulfuric acid react to produce potassium sulfate salt and water.
2. Potassium hydroxide and phosphoric acid react to produce potassium phosphate and water.
H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
3. Phosphoric acid sodium hydroxide react to produce sodium phosphate and water.
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
Answer:
KBr is limiting reactant.
Explanation:
Given data:
Mass of KBr =4g
Mass of Cl₂ = 6 g
Limiting reactant = ?
Solution:
Chemical equation:
2KBr + Cl₂ → 2KCl + Br₂
Number of moles of KBr:
Number of moles = mass/molar mass
Number of moles = 4 g/ 119 gmol
Number of moles = 0.03 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 6 g/ 70 gmol
Number of moles = 0.09 mol
Now we will compare the moles of reactant with product.
KBr : KCl
2 : 2
0.03 : 0.03
KBr : Br₂
2 : 1
0.03 : 1/2×0.03= 0.015
Cl₂ : KCl
1 : 2
0.09 : 2/1×0.09 = 0.18
Cl₂ : Br₂
1 : 1
0.09 : 0.09
Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.