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OlgaM077 [116]
3 years ago
7

Riley is testing how quickly sugar water freezes. He adds sugar water to one ice tray and plain water to another ice tray. He pl

aces each tray in the freezer and records the time when each one starts to freeze. What is the treatment in this experiment?
Water
Freezer
Ice tray
Sugar
Chemistry
2 answers:
trapecia [35]3 years ago
8 0

Answer:

The correct answer is sugar.

Explanation:

The sugar mentioned is the treatment in the given experiment. The plain water possesses a freezing point of zero degree Celsius. However, water treated with sugar or sugar water exhibits a lower freezing point in comparison to pure water. The treatment here, that is, the sugar makes lowering or depression of the freezing point of water. Water is the solvent and the sugar is the solute here.

qaws [65]3 years ago
4 0

The best and most correct answer among the choices provided by your question is the fourth choice.

Sugar is the dependent variable therefore,  it <span>is the treatment in the experiment.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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The protein catalase catalyzes the reaction The Malcolm Baldrige National Quality Award aims to:
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The question is missing a part, so the complete question is as follows:

The protein catalase catalyzes the reaction The Malcolm Bladrigde National Quality Awards aims to: 2H2O2 (aq) ⟶ 2H2O (l) + O2 (g) and has a Michaelis-Menten constant of KM = 25mM and a turnover number of 4.0 × 10 7 s -1. The total enzyme concentration is 0.012 μM and the intial substrate concentration is 5.14 μM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Calculate the initial rate, R (often written as V0), of this reaction.

1) Calculate Rmax

The turnover number (Kcat) is a ratio of how many molecules of substrate can be converted into product per catalytic site of a given concentration of enzyme per unit of time:

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where:

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Et is total enzyme concentration or concentration of total enzyme catalytic sites.

Calculating:

Kcat = \frac{Vmax}{Et}

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V0 = \frac{[S].(Vmax)}{KM + [S]}, where:

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7 0
3 years ago
What is the resulting pressure when a sample of gas at 25°C in a 500 ml balloon
Natali5045456 [20]

Answer:

233.56 torr

Explanation:

We'll begin by converting celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 25 °C

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Initial temperature (T₁) = 298 K

Final temperature (T₂) = 75 °C

Final temperature (T₂) = 75 °C + 273

Final temperature (T₂) = 348 K

Next, we shall convert 2 L to mL. This can be obtained as follow:

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Finally, we shall determine the resulting pressure. This can be obtained as follow:

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Initial volume (V₁) = 500 mL

Initial pressure (P₁) = 800 torr

Final temperature (T₂) = 348 K

Final volume (V₂) = 2000 mL

Final pressure (P₂) =?

P₁V₁/T₁ = P₂V₂/T₂

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Cross multiply

P₂ × 2000 × 298 = 400000 × 348

P₂ × 596000 = 139200000

Divide both side by 596000

P₂ = 139200000 / 596000

P₂ = 233.56 torr

Therefore, the resulting pressure is 233.56 torr

3 0
3 years ago
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