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hodyreva [135]
3 years ago
7

How do electrons differ from protons and neutrons? A. They do not move. B. They are larger. C. They are not in a fixed position.

D. They are located in the nucleus.
Chemistry
2 answers:
sergejj [24]3 years ago
6 0
Ian 890 is correct! :)
nikdorinn [45]3 years ago
4 0
Hi!

Electrons are particles which basically 'orbit' around the nucleus. Protons and neutrons are condensed, in a fixed position inside the nucleus. 

With this in mind, the answer will be C.

Hopefully, this helps! =)
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HELP ASAP 50-50 CHANCE
docker41 [41]

Answer: The answer is A, A new element or different atom formed from the original two.

Hope this helps! :D

-<em>TanqR</em>

4 0
3 years ago
A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

7 0
3 years ago
Molecule: Br2 and Br2. <br> Is it polar or nonpolar?
Ostrovityanka [42]
This combination in non polar.
7 0
3 years ago
How is nitric acid classified?
Darina [25.2K]
It's classified as an acid
4 0
3 years ago
How much heat was transferred from 50.0 g of water if the temperature of the water went from 30.0 ° C to 55.0 °? The specific he
Lena [83]

Answer:

Heat transfer = Q = 62341.6 J

Explanation:

Given data:

Heat transfer = ?

Mass of water = 50.0 g

Initial temperature = 30.0°C

Final temperature = 55.0°C

Specific heat capacity of water = 4.184 J/g.K

Solution:

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 = 55.0°C -  30.0°C

ΔT = 25°C (25+273= 298 K)

Q = 50.0 g × 4.184 J/g.K ×298 K

Q = 62341.6 J

4 0
3 years ago
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