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lora16 [44]
3 years ago
12

If two metals both have the same color, have similar densities, and are about the same size, but one is shiny and the other is d

ull, they can be separated based on the physical property known as what?
Chemistry
1 answer:
Papessa [141]3 years ago
7 0
<span>If two metals both have the same color, have similar densities, and are about the same size, but one is shiny and the other is dull, they can be separated based on the lustrous physical property of a metal. Lustrous is the property of the metal to shine/gleam gently because of its particle arrangement of atoms. </span>
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15 POINTS!!!
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<em>Answer:</em>

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As we know, the kinetic energy of substance is directly proportional to Temperature.

The total Kinetic energy can be calculated as follow

                                                K.E = 3/2 n.R.T

From the equations, it is cleared that Temperature is directly proportional to temperature.

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Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2
Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has SP^{2}  O has SP^{2} .

Explanation : The orbitals of oxygen and carbon which are involved in SP^{2} hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and  O also has π  orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b:  O SP^{3}  H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen SP^{3} hydrogen atoms in it. It has SP^{3} hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c:   C is SP^{3}  O is also SP^{3}

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of SP^{3} hybridized bonds.

Answer 5) Bond d:   C atom has SP^{3}  C atom has SP^{3}

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be SP^{3}. Therefore, the orbitals that undergo SP^{3} hybridization are SP^{3}.

Answer 6) Bond e : C1 containing O SP^{2}    

C2 is SP^{3}

Explanation : The carbon atom which contains oxygen along with a double bond has SP^{2} hybridized orbitals involved in the bonding process; whereas the carbon at C2 has SP^{3} hybridized orbitals involved during the bonding. This is for the 'e' position.

7 0
3 years ago
Consider an AB3 molecule in which A and B differ in elec-tronegativity. You are told that the molecule has an over-all dipole mo
Alborosie

Answer:

(b) Trigonal planar

Explanation:

The molecular geometry is the one that stabilizes better the bonds and the free electron pairs. If the molecule is nonpolar (overall dipole moment zero), so, there's no free electron pairs at the central atom. So, the molecule has the central atom A surrounded by three atoms of B, which is the trigonal planar geometry.

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3 years ago
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