As one moves across a period, from left to right, both the number of protons and electrons of a neutral atom increase. The enhancing number of electrons and protons results in a greater attraction between the electrons and the nucleus. This uplifted attraction pulls the electrons nearer to the nucleus, therefore, reducing the size of the atom.
On the other hand, while moving down a group, there is an increase in the number of energy levels. The enhanced number of energy levels increases the size of the atom in spite of the elevation in the number of protons. In the outermost energy levels, the protons get attracted towards the nucleus, however, the attraction is less due to an increase in the distance from the nucleus.
Because of the protons connected to the Nitrogen oxide group giving it its positive charge.
When calcium carbonate is heated, it breaks down to form calcium oxide and carbon dioxide.
Thermal decomposition is the process in which heat is required.
It is also known as thermolysis.
It is processed in which a compound breaks into two or more products when the heat is supplied.
This reaction is used for the production of oxygen.
This reaction is also used for production of acidic as well as basic oxides.
CaCO3 on thermal decomposition gives:
CaCO3→CaO+CO2
CaO→ Basic oxide.
CO2→ Acidic oxide.
Complete Question:
This diagram shows a marble with a mass of 3.8 grams (g) that was placed into 10 milliliters (mL) of water. Using the formula V M D = , what is the density of the marble?
(See attachment for full diagram)
Answer:
1.27 g/cm³
Explanation:
First, find the volume of rock:
Volume of rock = volume of water after rock was placed - volume of water before rock was placed
Volume of rock = 13 - 10 = 3ml
Density of rock = grams of rock per 1 cm³
Note: 1 ml = 1 cm³
Let x represent amount of rock per 1 cm³
Thus,
3.8g = 3 cm³
x = 1 cm³
Cross multiply
1*3.8 = 3*x
3.8 = 3x
3.8/3 = 3x/3
1.27 = x
Density of rock = 1.27 g/cm³
<span>- The tellurium isotopes that are most abundant have many neutrons found in their nuclei.</span>
