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3 years ago

Four students made a graphic organizer describing the parts of the atom.

Physics

2 answers:

Nikitich [7]3 years ago

8 0

Answer:

I believe the answer is D.

Explanation:

Protons are found inside the nucleus so are neutrons. Electrons are found outside the nucleus.

vodomira [7]3 years ago

5 0

Answer : The correct option is,

Particles Charge Location

Proton + inside nucleus

Electron - outside nucleus

Neutron 0 inside nucleus

Explanation :

As we know that an atom is the smallest unit of a matter that consist of three subatomic particles which are electrons, protons and neutrons.

The protons and the neutrons are located inside the nucleus or center of the nucleus where the mass of the an atom is concentrated and the electrons are located around the nucleus.

The protons are positively charged, the electrons are negatively charged and the neutrons are neutral that means it has no charge.

From the given table, we conclude that

The particle proton has positive charge and location is inside the nucleus.

The particle electron has negative charge and location is outside the nucleus.

The particle neutrons has no charge and location is inside the nucleus.

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You can see yourself in the mirror because?

jok3333 [9.3K]

Mirrors reflect light waves.

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3 years ago

Read 2 more answers

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago

1. Does the validity of the principle of conservation of momentum depend on the validity of Newton's 3rd law of motion?

OverLord2011 [107]

I truly believe so, that’s a definite yes

8 0

4 years ago

The second-order rate constants for the reaction of oxygen atoms ·with aromatic hydrocarbons have been measured (R. Atkinson and

Blizzard [7]

Answer: Frequency factor A = 8 × 10⁹

activation energy Ea = 15.5 KJ/Mol

Explanation: to begin, let us first define the parameters given;

K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹

K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹

K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹

also T₁ = 300.3 K

T₂ = 341.2 K

T₃ = 392.2 K

we know that;

㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]

where R is given as 8.314 J/mol-k

Ea = activation energy

K₁, K₂ = rate constant

T₁, T₂ = Temperature

therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]

this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol

∴ Ea = 15.5 KJ/ Mol

also given that K = A e⁻∧Ea/RT

here A = frequency factor

∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)

A = 7.99 × 10⁹ = 8 × 10⁹

3 0

3 years ago

How can the rate of a reaction be increased?

xeze [42]

I think it is D- diluting a solution sorry if i am wrong

7 0

3 years ago