Determine which decay process is being described in each case. the atomic number deceases by one, and the atomic mass is unchang
ed. the atomic number decreases by two, and the atomic mass decreases by four. the atomic number increases by one, and the atomic mass remains unchanged.
1. The atomic number deceases by one and the atomic mass is unchanged - β⁺/positron emission. In this a proton is converted to a neutron, hence no net change in mass. Since a proton is converted to neutron, the daughter nuclide's proton number /atomic number decreases by 1. Then atomic number decreases by 1 and no change in mass.
2.the atomic number decreases by two, and the atomic mass decreases by four - alpha decay / ⁴₂α. alpha particles have 2 neutrons and 2 protons. Since 2 protons and 2 neutrons are emitted, the mass of the nuclide decreases by 4. Since 2 protons have been emitted, this results in atomic number decreasing by 2.
3.<span>the atomic number increases by one, and the atomic mass remains unchanged. - </span>β⁻ beta decay. In this type of decay - beta decay , a neutron is converted to a proton, therefore no net change in mass. Since a proton is formed, atomic number increases by 1. Therefore atomic number increases by 1 and no change in mass.
<span>Ethylpropylamine has a chemical formula of C2H13N. It has a molecular weight of 83.166 g/mol. It's considered highly flammable and dangerous if swallowed. It should not come in contact with skin or in eyes and it should not be inhaled.</span>