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3 years ago

HELP How many electrons can exist in the second shell surrounding an atomic nucleus?

Physics

2 answers:

shepuryov [24]3 years ago

6 0

Aaaaaaaaaaaaaaaaaaaaaa hope it helps

svet-max [94.6K]3 years ago

3 0

More than 8 <span> electrons can exist in the second shell surrounding an atomic nucleus.</span>

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PLEASE DO ASAP IM POSTING ONE MORE

KIM [24]

Answer rain gauge measures rain shadow units millimetres

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2 years ago

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Darren filled ocean water, fresh water, bottled water, and tap water into four different containers. He then dropped identical g

elena-14-01-66 [18.8K]

I would say container 1

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2 years ago

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1) Momenum is

adelina 88 [10]

Answer:

momentum formula = Mass × Velocity

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2 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

2 years ago

Initially (at time t = 0) a particle is moving vertically at 7.5 m/s and horizontally at 0 m/s. Its horizontal acceleration is 1

Lelu [443]

Answer:

t = 0.657 s

Explanation:

given,

initial vertical velocity = 7.5 m/s

initial horizontal velocity = 0 m/s

angle = 49◦

using kinetic equation

final velocity in vertical direction

v sinθ = u_y - gt ........................(1)

final velocity in horizontal direction

v cosθ = u_x + a_x × t

here u_x = 0.0 m/s

v cosθ = a_x×t ......................(2)

Dividing equation (1) / (2)

$tan \theta =\dfrac{u_y - gt}{a_x\times t}$

solving for time t

$t = \dfrac{u_y}{tan \theta \times a_x + g}$

u_y = initial velocity along x direction

acceleration along a_x = 1.4 m/s²

g = acceleration due to gravity = 9.8 m/s²

θ = 43° , u_y = 7.5 m/s

$t = \dfrac{7.5}{tan 49^0\times 1.4+ 9.8}$

t = 0.657 s

time taken by the particle is t = 0.657 s

6 0

2 years ago