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1 month ago

Part 1 Designing an Investigation

1 answer:

3 0

The experimental setup involves assessing the temperature of the pizza, which serves as the dependent variable, after being allowed to cool in various thermal environments over a consistent time period used as a control. The following parameters are considered: The initial temperature of the pizza is 400°F, the freezer temperature is 0°F, the refrigerator is at 40°F, and the countertop is 78°F. The independent variable is the heat level experienced by the hot pizza, while the dependent one indicates the temperature it achieves during the cooling process. The plan for the experiment entails: 1) Positioning the pizza at 400°F in each heat setting (freezer, refrigerator, countertop) for the same duration, subsequently documenting the final temperature of the pizza. 2) The option yielding the lowest temperature after that timeframe indicates the fastest cooling method for the pizza.

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If Sara was planning a wedding and wanted to have a sculpture of a heart made of butter at the reception, describe how Sara coul

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49.9 g per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 300 m^3. P i

Alekssandra [3086]

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;

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= 2.079 g/hr

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Given the half-life = 3.4 hours

For a first-order reaction; k, the rate constant = ln2/t, where t is the half-time

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Substituting all parameters into the equation k $C_{A}$ V;

Mass lost to sunlight = k $C_{A}$ V

$C_{A}$ = Incoming mass per hour / kV

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0.034 g/m³ $C_{A}$

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1 month ago

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= $\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}$ . Since k has a value of 0.50357 at a temperature of 298 K, we can conclude that

$\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}$ = 0.50357, leading to $\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}$ + 1 = 1.50357, which equals $\frac{\text{[total fructose solution]}}{\text{[fructopyranose]}}$ $\frac{1}{1.50357}$ = 0.665. Therefore, we can deduce that 0.665 of the total fructose in the solution exists as fructopyranose.

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1 month ago