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Kruka [31]
2 years ago
9

Difference between shell and subshell?

Chemistry
1 answer:
juin [17]2 years ago
3 0
The "sub shells" are the orientations and shapes for your orbitals, going in order by Shells are a collection of subshells with the same principle quantum number, and subshells are a collection of orbitals with the same principle quantum number and angular momentum quantum number. Hope this helps :)
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Take mass of zinc divided by the relative molecular mass - 30 for zinc ( from periodic table)
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Semi-conductors have properties of both metals and non metals.which are two examples of metalloids
balu736 [363]
The six metalloids are boron, silicon, germanium, arsenic, antimony, and tellerium.
7 0
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What does the symbol "E3" represent?
Anna71 [15]

Answer:

your answer to your question is B

6 0
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Read 2 more answers
Anybody good at chemistry?
Luba_88 [7]

Answer:

Explanation:

1)

Given data:

Mass of lead = 25 g

Initial temperature = 40°C

Final temperature = 95°C

Cp = 0.0308 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 95°C -  40°C

ΔT = 55°C

Q = 25 g × 0.0308 j/g.°C  × 55°C

Q = 42.35 j

2)

Given data:

Mass  = 3.1 g

Initial temperature = 20°C

Final temperature = 100°C

Cp = 0.385 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 100°C -  20°C

ΔT = 80°C

Q = 3.1 g × 0.385 j/g.°C  × 80°C

Q = 95.48 j

3)

Given data:

Mass of Al = ?

Initial temperature = 60°C

Final temperature = 30°C

Cp = 0.897 j/g.°C

Heat released = 120 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 30°C -  60°C

ΔT = -30°C

120 j = m × 0.897 j/g.°C  × -30°C

120 j = m × -26.91  j/g

m = 120 j / -26.91  j/g

m =  4.46 g

negative sign show heat is released.

4)

Given data:

Mass of ice = 1.5 g

Change in temperature  = ?

Cp = 0.502 j/g.°C

Heat added= 30.0 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

30.0 j = 1.5 g × 0.502 j/g.°C  × ΔT

30.0 j = 0.753 j/°C  × ΔT

30.0 j /0.753 j/°C  = ΔT

39.84 °C  =  ΔT

3 0
3 years ago
For the reaction 2 Cr(s) + 3 Pb²⁺(aq) ⟶ 3 Pb(s) + 2 Cr³⁺(aq), what is the value of n in the Nernst equation?
Blizzard [7]

Answer:

x

Explanation:

xx

5 0
3 years ago
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