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tresset_1 [31]
3 years ago
15

What type of bond results from the side‑on overlap of orbitals?

Physics
1 answer:
Serga [27]3 years ago
4 0

Answer:

A pi bond

Explanation:

A pi bond is a type of covalent bond that results from the formation of a molecular orbital by the side-to-side overlap of atomic orbitals along a plane perpendicular to a line connecting the nuclei of the atoms.

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What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc
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The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

Q= mc\Delta \theta

where;

  • c is the specific heat capacity
  • Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC

Specific heat capacity of the metal for the third trial;

C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC

Specific heat capacity of the metal for the fourth trial;

C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC

Specific heat capacity of the metal for the fifth trial;

C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC

Average specific heat capacity

C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0
2 years ago
Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm.
pantera1 [17]

Answer:

R_a/R_b=3

Explanation:

The resistance in terms of the area and the length of the wire is given by:

R=pL/A

if we have two wires, the first one is a solid wire with a diameter of d_A = 1 x 10^-3 m, and the second one is a hollow wire with inner diameter of d_B,i = 1 x 10^-3 m and outer diameter of d_B,σ= 2 x 10^-3 m, so the cross sectional area of the first wire is:  

A_a=πr^2_a

A_a=πd^2_A/4

hence the resistance is:  

R_a=(4*p*L_a)/π*d^2_A                                     (1)

the area of the second wire is:  

A_b=π*r^2_B,σ-π*r^2_B,i

A_b=π/4(d^2_B,σ-d^2_B,i)

hence the resistance is:  

R_b=(4*p*L_b)/π(d^2_B,σ-d^2_B,i)                   (2)

To find the ratio between the resistances R_a/R_b, we divide (1) over (2) to get:  

R_a/R_b=(d^2_B,σ-d^2_B,i)*L_a/(d^2_a*L_b)

but the wires have the same length, therefore:  

R_a/R_b=(d^2_B,σ-d^2_B,i)/(d^2_a)

substitute with the given values to get:

R_a/R_b=3

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