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saveliy_v [14]
3 years ago
10

Determine whether each melting point observation corresponds to a pure sample of a single compound or to an impure sample with m

ultiple compounds.
Chemistry
1 answer:
daser333 [38]3 years ago
4 0

The question is incomplete, the complete question is;

Determine whether each melting point observation corresponds to a pure sample of a single compound or to an impure sample with multiple compounds.

Experimental melting point is BELOW literature value

Experimental melting point is CLOSE to literature value

WIDE melting point range

NARROW melting point range

Answer:

narrow melting point-pure sample of a single compound

experimental melting point is close to literature value-pure sample of a single compound

wide melting point range-impure sample of multiple compounds

experimental melting point is below literature value-impure sample of multiple compounds

Explanation:

The experimental melting point of a pure single compound is sharp and extremely close to the melting point of the substance as recorded in the literature. Usually, a pure substance melts within a narrow range of temperatures.

Impure samples of multiple compounds melt over a range of temperatures. Also if the experimental melting point is well below the record in literature, then the sample is contaminated by other compounds.

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Which of the following explains the process of radiation?
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Answer:

When heat gets transferred through electromagnetic waves that move through space .

Explanation:

Radiation is the propagation of energy in the form of electromagnetic waves or subatomic particles through a vacuum or a material medium.

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2 years ago
If the frequency of the photon increases what happens to the wavelength and energy?
kap26 [50]
The energy<span> per </span>photon<span> is proportional to the </span>frequency<span> of the radiation when considered as waves, ie inversely proportional to the </span>wavelength. Double the wavelength<span>, halve the </span>photon energy<span>. This means that long </span>wavelength<span> radiation (radio waves) has low </span>photon energy<span> and so does not penetrate matter.</span>
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3 years ago
2.What two factors must be held constant for density to be a constant ratio?
irina1246 [14]

Answer:

Temperature and Pressure

Explanation:

Temperature and pressure cause change in volume.

So any change in volume will alter the ratio of density as given by equation of density.

Density = mass/ volume

Change in volume will alter the ratio.

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6 0
2 years ago
Read 2 more answers
A solution is made by dissolving
barxatty [35]

Answer:

4.52 mol/kg

Explanation:

Given data:

Mass of lithium fluoride = 22.1 g

Mass of water = 188 g

Molality = ?

Solution:

Molality:

It is the number of moles of solute into kilogram of solvent.

Formula:

Molality = number of moles of solute / kilogram solvent

Mathematical expression:

m = n/kg

Now we will convert the grams of LiF into moles.

Number of moles = mass/ molar mass

Number of moles = 22.1 g/ 26 g/mol

Number of moles = 0.85 mol

Now we will convert the g of water into kg.

Mass of water = 188 g× 1kg/1000 g = 0.188 kg

Now we will put  the values in formula.

m = 0.85 mol / 0.188 kg

m = 4.52 mol/kg

8 0
3 years ago
How do you find the amount of moles is .032 grams of water and whats the answer
masha68 [24]

Answer:

\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

  • 2 Hydrogen atoms (subscript of 2)
  • 1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

  • Hydrogen: 1.0079 g/mol
  • Oxygen: 15.9994 g/mol

There are 2 hydrogen atoms, so multiply the mass by 2.

  • 2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

  • 2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

0.032 \ g  \ H_2 O* \frac{ 1 \ mol \ H_2 O}{18.0152 \ g \ H_2 O}

Multiply. The grams of H₂O will cancel out.

0.032 * \frac{1 \ mol \ H_2 O}{18.0152 }

\frac{0.032 *1 \ mol \ H_2 O}{18.0152 }

0.00177627781 \ mol \ H_2 O

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

0.0018 \ mol \ H_2 O

There are about <u>0.0018 moles in 0.032 grams.</u>

6 0
2 years ago
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