Answer:
Baking Soda is NaHCO3 ---Sodium hydrogen Carbonate
Excessive use of Fertilizer can cause plants to develop shallow root systems,Decrease soil fertility and reduce Quality of Crops.
Answer:
The ground state represents the lowest-energy state of the atom.
Explanation:
sd
Sugar increases the viscosity of water
hope that helps
Explanation:
from the graph study about oxygen content of Earth's atmosphere, we can understand that
1)
4 billions year ago = None, 3 billions year ago = Cyanobacteria and Archaea , 2 and 1 billions year ago = Bacteria and Green algae , 500 Ma = invertebrate fossils started to existence. Early land plants came in to existence around 398 MA that is Devonian. Dinosaurs are came in to existence during Triassic and Jurassic that is around 251 Ma. Man and animals are recent organism came under Holocene that is 11000 years ago.
2)
The first cells on the earth are anaerobic microorganisms, as the CO2 level is too high they survive by using CO2.
3)
Starting around 2.7 billion years ago, photosynthesis by Cyanobacteria and later plants , pumped “OXYGEN” in to the atmosphere. This caused the decline of anaerobic bacteria and allows the diversification of animals as seen in “CAMBRIAN” around 500 millions year ago.
Early vascular plants “CAPTURED” CO2 starting before the Carboniferous period that began around 350 millions year.Leading to lower temperatures and allowing and allowing the seed plants to outcompetes seedless plants.
Modern human activities has raised both “CO2 and METHANE” level in the atmosphere to over leading to higher temperature and extinction of other species.
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
