Answer:
Six
Explanation:
Mᵣ = 46
Aᵣ(C) = 0.522 × 46 = 24
Difference = 22
Less 1 O = -16
Aᵣ(H) = 6
The relative atomic mass of the H atoms is 6.
The molecule contains six hydrogen atoms.
CCl4
For central atom C
The formula to calculate hybridization is:
n= 1/2 [group no. central atom +no. of monovalent atoms attached to central metal atom (except oxygen) + magnitude of negative charge - magnitude of positive charge]
n= 1/2 [4 +4 + 0-0]
n= 1/2 [8)
=4
The hybridization is sp3 and the electron geometry is tetrahedral
A. All matter is composed of tiny particles that can vary in shape, size, and weight.
B. Solid matter is composed of square-shaped particles and liquid matter is composed of flat particles.
C. Solid matter is composed of atoms and liquid matter is composed of elements.
Answer:
PV=nRT
Explanation:
T=PV/nR
P=5.6atm n= 4 mol V=12 L R =.082Latm/K mol
T= (5.6 atm)(12 L)/(4 mol)(.082Latm/K mol)
T = 205 K = -68 C
An even distribution of electron density across the molecule, which means that the molecule has no distinct positive or negative ends. This lack of polarity makes the molecule symmetrical, and the forces of attraction between individual molecules are weak. Consequently, nonpolar molecules tend to have low boiling points and are generally insoluble in polar solvents.
When two oxygen atoms come together to form an oxygen molecule (O2), they share electrons to form a covalent bond. In this bond, each oxygen atom shares two electrons with the other oxygen atom, creating a stable molecule.
While it is true that the oxygen atoms in O2 molecule are electronegative and pull on the shared electrons in opposite directions, the two oxygen atoms are identical in terms of their electronegativity. This means that they pull on the electrons with equal strength, resulting in a balanced distribution of charge across the molecule. As a result, the molecule becomes nonpolar.
Click the below link, to learn more about Oxygen Atoms bond:
#SPJ11
A
B
D
AB
AO
O
Answer:
AB, A, O and B.
When 150 ml of 0.500 M silver nitrate are added to 100 mL of 0.400 M potassium chromate, a silver chromate precipitate forms. Considering the stoichiometry of the reaction and the quantities of reactants, 24.88 grams of silver chromate will precipitate.
The subject of this question is based on precipitation reactions in Chemistry. Precipitation reactions occur when two solutions combine to form an insoluble solid known as a precipitate. The moles of silver nitrate present in a 150 mL of 0.500 M solution can be calculated using the formula Molarity = Moles ÷ Volume (in Litres).
Thus, Moles of AgNO3 = 0.500 M * 0.15 L = 0.075 mol AgNO3. According to the reaction equation 2AgNO3 + K2CrO4 → 2AgCrO4(precipitate) + 2KNO3, for every mole of K2CrO4, we have two moles of AgNO3. Thus, based on stoichiometry and the given quantities of the reactants, the limiting reactant will be AgNO3, and it will totally react and form the silver chromate precipitate. The moles of Ag2CrO4 formed would therefore also be 0.075 mol. To convert this into grams, we use the molar mass of Ag2CrO4, which is approximately 331.73 g/mol. Hence, grams of Ag2CrO4 = 0.075 mol Ag2CrO4 * 331.73 g/mol = 24.88 g Ag2CrO4.
#SPJ12