Answer:
Kc = 9.52.
Explanation:
<em>A + 2B ⇌ C,</em>
Kc = [C]/[A][B]²,
Concentration: [A] [B] [C]
At start: 0.3 M 1.05 M 0.55 M
At equilibrium: 0.3 - x 1.05 - 2x 0.55 + x
0.14 M 1.05 - 2x 0.71 M
- For the concentration of [A]:
∵ 0.3 M - x = 0.14 M.
∴ x = 0.3 M - 0.14 M = 0.16 M.
∴ [B] at equilibrium = 1.05 - 2x = 1.05 M -2(0.16) = 0.73 M.
<em>∵ Kc = [C]/[A][B]²</em>
∴ Kc = (0.71)/(0.14)(0.73)² = 9.5166 ≅ 9.52.
If the ion has a charger of -1 and 10 electrons then it has 9 protons
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
B
Explanation:
A Lewis acid is specie that has vacant orbitals and is able to accept a lone pair of electrons while a Lewis base is any specie having a lone pair of electrons which can be donated to another chemical specie.
Hence a Lewis acid accepts a lone pair of electrons while a Lewis base donates a lone pair of electrons.
From the resonance structure shown, the carbon atom in the carbonyl compound ought to function as the Lewis acid since it is electrophilic (electron deficient).
1) during daylight, especially around noon, the relatively high air **temperature** and low humidity caused high evaporation, extracting pore water from the beach and leaving the **salt** behind, thereby resulting in high salinity near the beach surface (intertidal zone)
2) • algae and other intertidal plants grow in the abundant sunlight and support an entire food chain of animals
• constant wave action supplies the tide pool with **nutrients** and oxygen
• food is abundant
• a varied substrate provides hiding places and surfaces to cling to
3) the intertidal zone is one of a number of **marine** biomes or habitats, including: estuary (spray zone), neritic (lower/shallow zone), surface (middle zone), deep zones (high zones)
Explanation:
Rate constant = 6.08 x 10–4 s–1
What is the rate of the reaction when [N2O5] = 0.100 M..?
A first-order reaction depends on the concentration of one reactant, and the rate law is:
r= −dA / dt = k[A]
Rate of reaction = k [ A ]
where [A] = 0.100 M
Rate of reaction = (6.08 x 10–4) x 0.100
Rate of reaction = 6.08 x 10–5 Ms–1
what happens to the rate when the concentration of N2O5 is doubled to 0.200 M?
Rate of reaction = k [ A ]
[A] = 0.200
Rate of reaction = (6.08 x 10–4) x 0.200
Rate of reaction = 1.216 x 10–4 Ms–1
The rate of the reaction increases to 1.216 x 10–4 Ms–1