Answer:
All description is given in explanation.
Explanation:
Van der Waals forces:
It is the general term used to describe the attraction or repulsion between the molecules. Vander waals force consist of two types of forces:
1. London dispersion forces
2. Dipole-dipole forces
1. London dispersion forces:
These are the weakest intermolecular forces. These are the temporary because when the electrons of atoms come close together they create temporary dipole, one end of an atom where the electronic density is high is create negative pole while the other becomes positive . These forces are also called induce dipole- induce dipole interaction.
2. Dipole-dipole forces:
These are attractive forces , present between the molecules that are permanently polar. They are present between the positive end of one polar molecules and the negative end of the other polar molecule.
Hydrogen bonding:
It is the electrostatic attraction present between the atoms which are chemically bonded. The one atom is hydrogen while the other electronegative atoms are oxygen, nitrogen or flourine. This is weaker than covalent and ionic bond.
Ionic bond or electrostatic attraction:
It is the electrostatic attraction present between the oppositely charged ions. This is formed when an atom loses its electron and create positive charge and other atom accept its electron and create negative charge.
Hydrophobic interaction:
It is the interaction between the water and hydrophobic material. The hydrophobic materials are long chain carbon containing compound. These or insoluble in water.
Covalent bond:
These compounds are formed by the sharing of electrons between the atoms of same elements are between the different element's atoms. The covalent bond is less stronger than ionic bond so require less energy to break as compared to the energy require to break the ionic bond.
You add solute until it can't dissolve anymore.
Answer:
104.1 mL
Explanation:
c1V1 = c2V2
V1 = c2V2/c1 = (0.333 M×2500 mL)/8.00 M
V1 = 104.1 mL
Answer:
Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.
Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g × (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g × (1 mole/ 32 g) = 1.53 moles O2 B.
Explanation:
MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield
3 these are 5,6&7 are the significant
