Answer:
5.645 × 10⁻²³ g
Solution:
Step 1) Calculate Molar Mass of SH₂;
Atomic Mass of Sulfur = 32 g/mol
Atomic Mass of H₂ = 2 g/mol
--------------------
Molecular Mass of SH₂ = 34 g/mol
Step 2: Calculate mass of one molecule of SH₂ as;
As,
Moles = # of Molecules / 6.022 × 10²³
Also, Moles = Mass / M.Mass So,
Mass/M.mass = # of Molecules / 6.022 × 10²³
Solving for Mass,
Mass = # of Molecules × M.mass / 6.022 × 10²³
Putting values,
Mass = (1 Molecule × 34 g.mol⁻¹) ÷ 6.022 × 10²³
Mass = 5.645 × 10⁻²³ g
Answer:
The lanthanides and the actinides at the bottom of the table are sometimes known as the inner transition metals because they have atomic numbers that fall between the first and second elements in the last two rows of the transition metals
, a crystal structure with a short symmetrical hydrogen bond.
<h3>What is Classical bonding?</h3>
Classical models of the chemical bond. By classical, we mean models that do not take into account the quantum behaviour of small particles, notably the electron. These models generally assume that electrons and ions behave as point charges which attract and repel according to the laws of electrostatics.
Sodium dihydrogen phosphate is a derivative composed of glycerol derivatives formed by reacting mono and diglycerides that are derived from edible sources with phosphorus pentoxide followed by neutralization with sodium carbonate.
Bonding in
, a crystal structure with a short symmetrical hydrogen bond. Sodium dihydrogen phosphate () is monoclinic, space group P2,/c, with a= 6.808 (2), b= 13.491 (3), c=7.331 (2)/~, fl=92.88 (3) ; Z=8.
Learn more about the bond here:
brainly.com/question/10777799
#SPJ1
Answer:
Explanation:
The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds.
The term oxidation was first used by Antoine Lavoisier to signify reaction of a substance with oxygen. Much later, it was realized that the substance, upon being oxidized, loses electrons, and the meaning was extended to include other reactions in which electrons are lost, regardless of whether oxygen was involved.
Helped?
Brainliest?
Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
