Although I may not be the smartest, I can definitely answer.
This represents a chemical change because the substances' chemical identities were altered. The fizzing was a clear sign, and the temperature increase was another indicator of the reaction.
1) To express 0.89% m/v, it equals 0.89 grams of NaCl per 100 ml of solution.
This corresponds to 8.9 grams of NaCl in 1000 ml of solution, or 8.9 grams in 1 liter.
2) Molarity is represented as M = moles of solute / liters of solution.
Thus, we need to determine the moles in 8.9 grams of NaCl.
3) The molar mass of NaCl is calculated as 23.0 g/mol + 35.5 g/mol = 58.5 g/mol.
4) Therefore, the number of moles of NaCl calculates as mass / molar mass = 8.9 g / 58.5 g/mol = 0.152 moles.
5) Consequently, M = 0.152 moles of NaCl / 1 liter of solution = 0.152 M.
Answer: 0.152 M
The inquiry is incomplete; here is the full question:
One tank of goldfish receives the standard amount of feeding once daily, a second tank is given two feedings a day, and a third tank is fed four times daily throughout a six-week experiment. The body fat of the fish is recorded every day.
Independent Variable-
Dependent Variable-
Constants
Control Group-
Answer:
A) The quantity of food given to the goldfish
B) The body fat of the goldfish
C) -Type of fish in the experiment (goldfish)
Time period for feeding the fish (six weeks)
Shape and size of the tanks
D) group of goldfish receiving the standard feeding amount
Explanation:
The objective of the experiment is to assess how the quantity of food affects the body fat of goldfish. Consequently, the amount of food serves as the independent variable while the body fat acts as the dependent variable.
The control group is the one given the standard feeding amount (once daily). All subjects are goldfish, fed over a six-week duration, with all tanks being the same shape and size, establishing the constants in the research.
The pH level is 1.39. To explain, we start with the given information: the concentration of HClO is 0.15 M, with an acid dissociation constant of 2.9 × 10-8. The objective is to calculate the pH of the solution. Through the process, we find that the equilibrium concentration after applying the formula yields 0.04069 M for H3O⁺, leading us to a pH of 1.39.
