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Nina [5.8K]
4 years ago
10

A globe is placed on your desk and you are asked to identify its physical properties. How would you describe the globe?

Chemistry
1 answer:
gogolik [260]4 years ago
8 0
You could describe it's shape, color, varieties on sizes and shape of states, how it's rounded but with a top and bottom.
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In 1860, Chemists could make which of the following statements about the known chemical elements?
LenKa [72]

Answer:

b. Some had similar properties

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3 years ago
What is the type of weak bond between the hydrogen of one molecule and the nitrogen of another molecule, where the two don't act
Aliun [14]
<h2>Hydrogen Bonding</h2>

Explanation:

  • Hydrogen bonding is a type of weak bond.
  • Hydrogen Bonding occurs when the hydrogen atom of one molecule is bonded with a molecule of high electronegativity element like nitrogen of another molecule.
  • It is a type of dipole-dipole interations.
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  • Hydrogen bonds are weaker than covalent and ionic bonds.
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7 0
3 years ago
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In order to reduce the exposure to organic solvents like turpentine, some art instructors recommend the students clean brushes a
kykrilka [37]
Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.

To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.
6 0
3 years ago
To what Kelvin temperature must a balloon
snow_tiger [21]

Answer:

T_2=261.46\ K

Explanation:

It is given that,

Original temperature, T_1=323^{\circ}C=596.15\ K

Original volume, V_1=2.85\ L

We need to find the temperature if the volume of the balloon to be shrink to 1.25 L.

According to Charles law, at constant pressure, V\propto T

It would means, \dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}

T₂ = ?

T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{1.25\times 596.15}{2.85}\\\\T_2=261.46\ K

So, the new temperature is 261.46 K.

4 0
3 years ago
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