A covalent bond which is also called as molecular bond, is a chemical bond which involves mainly the sharing of electron pairs between atoms. These electron pairs are called as shared pairs or bonding pairs, which are stable balance of attractive and repulsive forces between atoms, when these bonding pairs share electrons, it is called as covalent bonding.

Covalent bonding is usually seen in non-metals which share their electrons while bonding.

4 0

3 years ago

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Rate = K[A]

Aleks [24]

Answer:

9 × 10⁻³ mol·L⁻¹s⁻¹

Explanation:

Data:

k = 1 × 10⁻³ L·mol⁻¹s⁻¹

[A] = 3 mol·L⁻¹

Calculation:

rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹

8 0

3 years ago

Which type of system must exist for chemical equilibrium

larisa86 [58]

Answer: closed system.

Explanation:

4 0

3 years ago

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wave

timofeeve [1]

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

$n_i=8$

$\lambda=3745\ nm$

The amount of energy change during the transition is given by :

$\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]$

And

$\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]$

Plugging all the values we get :

$\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5$

So, the final level of the electron is 5.

4 0

3 years ago