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Nata [24]
3 years ago
15

A small amount of a solid unknown compound was placed into a test tube containing isopropanol. The solid floats on top of the li

quid isopropanol. Is this unknown compound considered to be soluble or insoluble?
Chemistry
2 answers:
Marat540 [252]3 years ago
7 0

Answer:

It is insoluble

Explanation:

Solubility and floatability are different concepts:

Solubility depends of the polarity and chemical characteristics of the solute and solvent.  

Floatability depends of the density and volume of the solid, and the density of the liquid it's in.

Given that the unknown solid didn't disolve (you can see it on top of the liquid), it is insoluble in isopropanol.

masya89 [10]3 years ago
3 0
The unknown solid did not dissolve into the isopropanol , therefore it is insoluble. 
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How many moles of calcium (Ca) atoms are there in 77.4 g of Ca?
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Describe a mechanical property of steel that characterizes its resistance to permanent deformation
strojnjashka [21]

The mechanical property of steel that characterizes its resistance to permanent deformation is ductility.

Am alloy is a substance prepared by adding one or more elements to a base or parent metal to obtain desirable properties.

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The mechanical or physical properties of steel include:

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Therefore, the mechanical property of steel that characterizes its resistance to permanent deformation is ductility.

Learn more here:

brainly.com/question/21706983

8 0
3 years ago
The decomposition of kclo3 proceeds spontaneously when it is heated. do you think that the reverse reaction, the formation of kc
pishuonlain [190]

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7 0
4 years ago
Sulfur exists in many forms with the general molecular formula Sn. If 1.00 g of sulfur is added to 500.0 g of carbon tetrachlori
Ber [7]

Answer:

S₁₂

Explanation:

The freezing point depression (ΔTf) is a colligative property that can be calculated using the following expression.

ΔTf = Kf × m

where,

Kf: freezing point depression

m: molality

ΔTf = Kf × m

m = ΔTf / Kf

m = 0.156 °C / (29.8 °C/m)

m = 5.23 × 10⁻³ m

The molality is:

m = moles of solute / kilograms of solvent

moles of solute = m × kilograms of solvent

moles of solute = 5.23 × 10⁻³ mol/kg × 0.5000 kg

moles of solute = 2.62 × 10⁻³ mol

1.00 g corresponds to 2.62 × 10⁻³ moles. The molar mass of Sₙ is:

1.00 g/2.62 × 10⁻³ mol = 382 g/mol

We can calculate n.

n = molar mass of Sₙ / molar mass of S

n = (382 g/mol) / (32.0 g/mol)

n = 11.9 ≈ 12

The molar formula is S₁₂.

4 0
3 years ago
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