Answer:
2.803013439419911 × 10⁻¹² J
Explanation:
Mass defect = mass of reactant - mass of product
(2.0140 + 3.01605) - (4.002603 + 1.008665)
5.03005 - 5.011268 = 0.018782 amu
mass in Kg = mass (amu) × 1.66053892173 × 10⁻²⁷ kg
mass in kg = 0.018782 × 1.66053892173 × 10⁻²⁷ = 3.1188242027932 × 10⁻²⁹kg
E = Δm c² where c is the speed of light = 2.9979 × 10⁸m/s
E = 3.1188242027932 × 10⁻²⁹kg × (2.9979 × 10⁸m/s)² = 2.803013439419911 × 10⁻¹² J
Answer:
period 6
Explanation:
If the lanthanides were inserted into the periodic table, they would go into periodic 6.
Their atomic number is between 57 - 71 from element lanthanum to lutetium.
- The elements in this period are 15 in number.
- They are also know as elements in the f-block.
The elements that makes up the series are:
Lanthanum
Cerium
Praseodymium
Neodymium
Promethium
Samarium
Europium
Gadolinium
Terbium
Dysprosium
Holmium
Erbium
Thulium
Ytterbium
Lutetium
Answer:
The number of neutrons present in one atom of isotope of Silicon of mass 28 amu is<u> 14 neutrons</u>
Explanation:
Symbol of Si isotope
<u>Number of Neutron = Mass number - Atomic Number</u>
Mass number = Total number of protons and neutrons present in the nucleus of the atom.For Si = 28 amu
Atomic Number = Total number of Protons present in the nucleus.
Si = 14
Number of neutron = 24 - 14
= 14
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