Answer:
Mutualisms/Symbiotic relationships
Explanation:
It’s a mutual beneficial relationship that helps both sides… for example… clownfish and sea anemones. The sea anemones provide the clownfish protection and shelter, and the clownfish provides nutrients from waste for the anemones.
Hope that helped!
In a chemical reaction, the equilibrium constant refers to the value of its reaction quotient at chemical equilibrium, that is, a condition attained by a dynamic chemical system after adequate time has passed, and at which its composition has no measurable capacity to undergo any kind of further modification.
The given reaction is: HCN (aq) + OH⁻ = CN⁻ (aq) + H2O (l)
The equilibrium constant = product of concentration of products / product of concentration of reactants
(Here, H2O is not considered as its concentration is very high)
So, Keq = [CN⁻] / [HCN] [OH⁻]
Answer:
c is the answer then check it out
Answer:
Option D is correct = 8.12 grams of NaCl
Explanation:
Given data:
Moles of sodium chloride = 0.14 mol
Mass of sodium chloride = ?
Solution:
Formula:
Number of moles = mass of NaCl / Molar mass of NaCl
Molar mass of NaCl = 58 g/mol
Now we will put the values in formula.
0.14 mol = Mass of NaCl / 58 g/mol
Mass of NaCl = 0.14 mol × 58 g/mol
Mass of NaCl = 8.12 g of NaCl
Thus, 0.14 moles of NaCl contain 8.12 g of NaCl.
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
