Answer:
15.71g
Explanation:
The combustion equation that applies to hydrocarbons is
CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O
In the case of octane, C8H18:
C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O
C8H18 + 50/4 O2 = 8CO2 + 9H2O
C8H18 + 25/2 O2 = 8CO2 + 9H2O
2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)
From this balanced reaction,
2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide
That means,
44.8 L of octane generates 704g of carbon dioxide
Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide
Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.
Answer:
A, B, and C
Explanation:
Indeed, atoms possess mass and serve as the fundamental building blocks of chemical elements. While matter is composed of atoms, these particles themselves do not occupy physical space.
Atoms consist mostly of void, which excludes them from the other responses.
This confirms that A, B, and C are the right choices.
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.
Hi there! Calvin informed Marie that they could still incorporate solute until reaching 40 grams because the solution remained unsaturated. Unsaturated solutions denote situations where the solvent (water in this instance) can further dissolve more solute (here, KNO₃) considering the current pressure and temperature. This can be visually confirmed when additional solute does not lead to visible solid residues settling at the bottom of the flask, indicating that the dissolving rate surpasses the crystallization rate. Wishing you a pleasant day!
First, we need to identify the half-reaction for magnesium. It can be represented as:
Mg2+ + 2e- = Mg
Next, we will determine the overall charge generated during the electrolysis using the information derived from the half-reaction. The calculation follows:
4.50 kg Mg (1000 g / 1 kg) (1 mol / 24.305 g) (2 mol e- / 1 mol Mg) (96500 C / 1 mol e-) = 35733388.2 C
The provided EMF is given in voltage. Since 1 V equals J/C, 5 V translates to 5 J/C.
Therefore, 35733388.2 C (5 J/C) = 178666941 J
Finally, 178666941 J (1 kW-h / 3.6x10^6 J) = 49.63 kW-h
