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Rudik [331]
3 years ago
9

How could you predict the sinking/floating behavior of a spherical object?

Chemistry
2 answers:
antiseptic1488 [7]3 years ago
8 0
You can predict it by measuring the weight of the spherical object and measuring the liquid (mL) and subtracting finding the volume
worty [1.4K]3 years ago
3 0

To predict the sinking/floating behavior of a spherical object, you could determine its density.

<em>Step 1</em>. Determine its <em>mass</em> on a suitable balance.

<em>Step 2</em>. Determine its <em>volume</em>

<em>Either </em>

Measure its circumference. <em>C</em> = 2π<em>r</em>, so you can calculate r

<em>or </em>

Measure its diameter. <em>d</em> = 2<em>r</em>, so you can calculate <em>r </em>

<em>Then </em>

<em>V</em> = ⁴/₃π<em>r</em>³ , so you can calculate its volume.

<em>Step 3</em>. Calculate its <em>density </em>(<em>D</em>).

Density = mass/volume

If D < 1 g/cm³, the object will float in water. If D > 1 g/cm³, the object will sink.

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Enter a molecular equation for the gas-evolution reaction that occurs when aqueous hydroiodic acid and aqueous potassium sulfite
Katen [24]

Answer:

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3 years ago
Org. chem. How is this "1-aminopropan-2-ol" and not "3-aminopropan-2-ol"
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2 years ago
Which chemical equation describes the balanced reaction between lead and oxygen to form lead oxide?
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7 0
3 years ago
Read 2 more answers
What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

<u>Given:</u>

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

<u>To determine:</u>

The mean free path, λ

<u>Calculation:</u>

The mean free path is related to the collision cross section by the following equation:

\lambda =\frac{1}{n\sigma }------(1)

where n = number density

n = \frac{P}{kT}-----(2)

Substituting for P, k and T in equation (2) gives:

n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\  m^{3}

Next, substituting for n and σ in equation (1) gives:

\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m

6 0
3 years ago
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