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ra1l [238]
3 years ago
10

Students were investigating properties of matter, but wanted to make sure that when they tested these properties they had proper

ties that would represent the matter they were testing specifically. The students claim that they are going to test intrinsic properties. What evidence and reasoning would you provide to justify this claim? (7 points)
Chemistry
1 answer:
Mariana [72]3 years ago
4 0

Answer:

See explanation

Explanation:

An intrinsic property is a property that is internal, that is, it characterizes the substance under study. The possession of an intrinsic property depends on the nature of the substance. An intrinsic property does not depend on amount of substance but on the nature of the substance.

Examples of intrinsic properties include; Density. Solubility, Melting Point, Freezing Point, Boiling Point, Conductivity etc.

Intrinsic properties really represent the matter that is being studied. For instance, the boiling point of water will always be 100°c. No other liquid can boil exactly at that temperature. Hence, this intrinsic property can always be used to identify an unknown liquid as water.

The students were right, studying intrinsic properties accurately represent the matter that is being studied.

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If gas is $4.00/gal, how much does it cost to travel from Portland, Maine, to Orlando, Florida (a distance of 1,500 mi)? Give yo
AlexFokin [52]

it cost $200 , the answer is not $6000

6 0
3 years ago
Read 2 more answers
On the basis of chemical structure, which solid will most easily conduct electricity?
Marizza181 [45]

Answer:

Fe

Explanation:

The electrical conductivity depends mainly on the type of chemical bonds between the atoms of a compound.

In the case of MgF2, FeCl3 and FeO3, these have the type of ionic bond. This means that in the atoms of the compound there is an electron transfer, to keep eight electrons in the outermost layer and thus resemble the electronic configuration of the inert gas closest to each of the two elements, due to this ions of opposite charges are formed that are held together by electrostatic forces. These types of compounds are good conductors of electricity, however, to have this property, they must be dissolved in water or molten.

In the case of Fe, however, the type of union between atoms is metallic. In this type of junction, valence electrons are quite free inside the metal, which makes it easy for them to move. For this reason, this compound will conduct electricity in a solid state.

6 0
3 years ago
If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give
tankabanditka [31]

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

  • The maximum height attained by nitrogen gas molecule = h
  • The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

  • The potential energy at height h = P.E = 6.381\times 10^{-21} J
  • Molar mass of nitrogen gas =  28.0134 g/mol
  • Mass of nitrogen gas molecule = m

m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg

  • The acceleration due to gravity = g = 9.8 m/s^2
  • The maximum height attained by nitrogen gas molecule = h
  • The potential energy is given by:

P.E=mgh

6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Learn more about the average kinetic energy of gas particles here:

brainly.com/question/16615446?referrer=searchResults

brainly.com/question/6329137?referrer=searchResults

6 0
2 years ago
Please help me with this chem question, I’ll mark you brainiest if it’s right. There’s multiple answers for this one.
Kazeer [188]

Answer:

Option A. KCl (aq)

Option D. Mg(OH)₂(s

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

MgCl₂(aq) + KOH(aq) —>

In solution, MgCl₂(aq) and KOH(aq) will dissociate as follow:

MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)

KOH(aq) —> K⁺(aq) + OH¯(aq)

MgCl₂(aq) + KOH(aq) —>

Mg²⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + OH¯(aq) —> 2K⁺(aq) + 2Cl¯(aq) + Mg(OH)₂ (s)

MgCl₂(aq) + KOH(aq) —> 2KCl (aq) + Mg(OH)₂(s)

Thus, the products of the above reaction are: KCl(aq) and Mg(OH)₂(s)

Thus, option A and D gives the correct answer to the question.

8 0
2 years ago
For the chemical reaction
polet [3.4K]

Answer:

806.3g

Explanation:

Given parameters:

 Number of moles of silver nitrate  = 4.85mol

Unknown:

Mass of silver chromate = ?

Solution:

         2AgNO₃ + Na₂CrO₄  →  Ag₂CrO₄ + 2NaNO₃

To solve this problem, we work from the known to the unknown;

  • The known specie here is  AgNO₃ ;

   From the balanced chemical equation;

          2 moles of AgNO₃  will produce 1 mole of Ag₂CrO₄

          4.85 moles of AgNO₃  will produce \frac{4.85}{2}   = 2.43moles of Ag₂CrO₄

  • Mass of silver chromate produced;

        mass = number of moles x molar mass

   Molar mass of  Ag₂CrO₄

    Atomic mass of Ag = 107.9g/mol

                                 Cr  = 52g/mol

                                  O  = 16g/mol

  Input the parameters and solve;

     Molar mass  = 2(107.9) + 52 + 4(16) = 331.8g/mol

  So,

        Mass of Ag₂CrO₄ = 2.43 x 331.8 = 806.3g

     

8 0
2 years ago
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