Convert 55.0g Ca(OH)2 to moles.
The calculation shows that 55.0g of Ca(OH)2 corresponds to 0.742 moles.
To find the volume, divide 0.742 mol of Ca(OH)2 by its molarity of 0.680M, yielding approximately 1.09L of Ca(OH)2.
If you disregard the negligible volume of the Ca(OH)2 itself, the resulting total volume of a 0.680M solution created by dissolving 55.0g of Ca(OH)2 in an appropriate amount of water would be 1.09L.
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.
Answer:
The concentration of P in the pond at equilibrium is 0.034 g/m³
Explanation:
Given the total mass = 49.9 g
1 day = 24 hours
mass per hour;
Incoming mass = (49.9 g / day) * (1 day /24 hr
)
= 2.079 g/hr
Outgoing mass = 0
Mass lost due to sunlight = k
V
Given the half-life = 3.4 hours
For a first-order reaction; k, the rate constant = ln2/t, where t is the half-time
ln 2= 0.693, V= volume
k = 0.693 / t_half = 0.693 / 3.4 = 0.2038 hr⁻¹
Substituting all parameters into the equation k
V;
Mass lost to sunlight = k V
= Incoming mass per hour / kV
= 2.079 g/hr / (0.2038 hr⁻¹ x 300 m³)
=
0.034 g/m³
Molarity is defined as the number of moles present in one liter of solution. Given the mass of NH₃ is 2.35 g and its molar mass is 17 g/mol, the moles of NH₃ in 2.35 g can be calculated as 2.35 g / 17 g/mol = 0.138 mol. Consequently, in a 0.05 L solution, the number of moles amounts to 0.138 mol. Therefore, the concentration in 1 L is: 0.138 mol / 0.05 L x 1L = 2.76 mol. Thus, the molarity of NH₃ is 2.76 M.
