First of all, there are five types of solid materials:
Metallic solids which are solids composed of metal atoms that are held together by metallic bonds.
Network solid is a chemical compound in which the atoms are bonded by covalent bonds in a continuous network extending throughout the material.
Molecular solid is a solid consisting of discrete molecules.
Ionic solid is a chemical compound composed of ions held together by electrostatic forces termed ionic bonding.
Amorphous solid is non-crystalline solid that lacks the long-range order that is characteristic of a crystal.
Now, after the defined all the types of solid materials in the equation lets to solve it.
A. the answer is the network solids, because covalent bonds are relatively strong, covalent are typically characterized by hardness, strength, and high melting points.
B. the answer is the metallic solids, due to that heat conduction occurs when a substance is heated and the particles will gain more energy vibrating more. These molecules then bump into nearby particles and transfer some of their energy to them and in metals this process have a higher probability than in the case of other solids due to the nature of the chemical bonds. It also has a range of hardness due to the strength of metallic bonds which varies dramatically.
C. the answer is the ionic solid; due to positive and negative ions which are bonded to form a crystalline solid held together by charge attractions.
To find the molar mass<span> of </span>Ba(NO3)2<span>, determine the </span>molar masses of all the atoms that form it. The Molar mass for Barium nitrate is <span>261.337 g/mol.</span>
<span>You are given a cough syrup that contains 5.0% ethyl alcohol, c2h5oh, by mass and its density of the solution is 0.9928 g/ml. The molarity of the alcohol in the cough syrup is 21.55.</span>
The balanced equation for the above reaction is as follows;
<span>10Na(s) + 2KNO</span>₃(s)--> K₂O(s) + 5Na₂O(s) + N₂<span>(g)
Stoichiometry of Na to KNO</span>₃ is 10 : 2
Number of Na moles reacted - 5.1 g/ 23 g/mol = 0.22 mol
Na is the limiting reactant, therefore Na is fully used up, Since KNO₃ is present in excess , at the end of the reaction a certain amount of KNO₃ will be remaining.
10 mol of Na reacts with 2 mol of KNO₃
Therefore 0.22 mol of Na reacts with - 2 /10 x 0.22 = 0.044 mol of KNO₃
Mass of KNO₃ reacted = 0.044 mol x 101.1 g/mol = 4.45 g
Mass of KNO₃ present initially - 305 g
Therefore remaining mass of KNO₃ - 305 - 4.45 = 300.55 g