Every organic molecules/compound contains carbon (c).
Some other very abundant are hydrogen, nitrogen, oxygen, phosphorus, and sulfur.
I learned this with the acronym CHNOPS.
C - Carbon
H - Hydrogen
N - Nitrogen
O - Oxygen
P - Phosphorus
S - Sulfur
Hope this helps!
Answer:
82.0 mL
Explanation:
Step 1: Given data
- Concentration of concentrated acid (C₁): 12.2 M
- Volume of concentrated acid (V₁): ?
- Concentration of dilute acid (C₂): 1.00 M
- Volume of dilute acid (V₂): 1.00 L
Step 2: Calculate the required volume of the concentrated acid
We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the concentrated acid using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 1.00 M × 1.00 L / 12.2 M = 0.0820 L = 82.0 mL
In a balloon there is a low pressure within the balloon. This means that the particles are more spaced out. Wheras in the cylinder they are under high pressure, meaning that the particles are much closer together!
The single most important chemical weathering agent is Carbon dioxide.
Weathering refers to the process that change the physical and chemical character of rock at or near the surface. Weathering has a dramatic impact on the composition of Earth's atmosphere. Chemical weathering removes carbon dioxide from the atmosphere, allowing it to be transformed into limestone and stored in the crust. Without chemical weathering, the elevated levels of carbon dioxide in the atmosphere would have long made Earth too hot to sustain life.
Answer:
ΔG° = -533.64 kJ
Explanation:
Let's consider the following reaction.
Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)
The standard Gibbs free energy (ΔG°) can be calculated using the following expression:
ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)
where,
ni are the moles of reactants and products
ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products
ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)
ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)
ΔG° = -533.64 kJ