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ra1l [238]
3 years ago
15

Calculate the approximate volume of a 1.50 mol sample of gas at 15.0°C and a pressure of 3.75 atm

Chemistry
1 answer:
Papessa [141]3 years ago
8 0

Answer:

The answer is J (9.46 L)

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describe in general terms an experiment to determine the molal freezing point depression constant kf of water. Assume the availa
Dvinal [7]
A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point: 
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b -  molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.

6 0
3 years ago
Zinc, sulphur, oxygen<br> What would this compound be called?
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This is the answer.

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3 years ago
Thinking and questioning is the start of the scientific inquiry process. Please select the best answer from the choices provided
aev [14]
Yes thats true! You always have to think about the question or project before you start a science experiment! :) 




8 0
3 years ago
Read 2 more answers
What does the 3 indicate in 1s22s22p63s1?
Dahasolnce [82]
The 3 indicates the third electron shell. (Which has only 1 electron in it in this configuration)

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3 years ago
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A rubber balloon was filled with helium at 25.0˚C and placed in a beaker of liquid nitrogen at -196.0˚C. The volume of the cold
Ksenya-84 [330]

Answer:

The volume of helium at 25.0 °C is 60.3 cm³.

Explanation:

In order to work with ideal gases we need to consider absolute temperatures (Kelvin). To convert Celsius to Kelvin we use the following expression:

K = °C + 273.15

The initial and final temperatures are:

T₁ = 25.0 + 273.15 = 298.2 K

T₂ = -196.0 + 273.15 = 77.2 K

The volume at 77.2 K is V₂ = 15.6 cm³. To calculate V₁ in isobaric conditions we can use Charle's Law.

\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}} \\V_{1}=\frac{V_{2}}{T_{2}} \times T_{1}=\frac{15.6cm^{3} }{77.2K} \times 298.2K=60.3cm^{3}

3 0
3 years ago
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