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3 years ago

Classify the following substances as a Bronsted-Lowry acid, Bronsted-Lowry base, Lewis acid, and/or Lewis base.

Chemistry

1 answer:

Margarita [4]3 years ago

7 0

Answer :

HCl : Bronsted-Lowry acid

$BF_3\text{ and }^{+}CCl_3$ : Lewis-acid

$^-H, CH_2O,^-OCH_3\text{ and }NH_3$ : Bronsted-Lowry base

$^-H, CH_2O,^-OCH_3,CH_3Cl\text{ and }NH_3$ : Lewis-base

Explanation :

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

HCl :

It is a Bronsted-Lowry acid that donates protons.

$BF_3\text{ and }^{+}CCl_3$ :

It is a Lewis-acid that accepts electron pairs.

$^-H, CH_2O,^-OCH_3\text{ and }NH_3$ :

It is a Bronsted-Lowry base that accepts protons.

$^-H, CH_2O,^-OCH_3,CH_3Cl\text{ and }NH_3$ :

It is a Lewis-base that donates electron pairs.

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3 years ago

Choose the options below that are true of a solution of a solid in a liquid. (select all that apply) Select all that apply: Most

Alekssandra [29.7K]

Answer:

Most solids in solution exhibit a general trend of increasing solubility with increasing temperature.

A seed crystal may be added to a supersaturated solution to precipitate excess solute.

Explanation:

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3 years ago

3) Explain why elements in the same group<br>similar properties

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Answer:

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3 years ago

Helppp ! Please :’)

Over [174]

Answer:

$\boxed{\sf \lambda \: = 1.282 \times {10}^{ 3} nm}$

Explanation:

Suppose an electron makes transition from n initial (ni) to n final (nf) then formula for wavelength is given by,

$\frac{1}{ \lambda} = R_H {Z}^{2} \big(\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}} \big)$

Where,

λ is wavelength of photon
Rʜ is rydberg constant, the value of Rʜ is

109690 Cm-¹ in <em>Puri Sharma Pathania</em> standard book of physical chemistry &

109737 Cm-¹ according to <em>Wikipedia</em>

Z is the atomic number of atom, for hydrogen Z =1,

& according to given data, ni = 5, nf = 3

Solution:

Let's solve for wavelength,

Substituting all the given data in above formula,

$\frac{1}{ \lambda} = 109690 \times {1}^{2} \big(\frac{1}{{3^2}} - \frac{1}{{5^2}} \big)$

$\frac{1}{ \lambda} = 109690 \times {1}^{2} \big(\frac{1}{{9}} - \frac{1}{{25}} \big)$

$\frac{1}{ \lambda} = 109690 \times {1}^{2} \times \frac{25 - 9}{25 \times 9}$

$\frac{1}{ \lambda} = 109690 \times {1}^{2} \times \frac{16}{225}$

$\frac{1}{ \lambda} = 7800.18 \: \: cm ^{ - 1}$

$\lambda = 1.282 \times {10}^{ - 4} cm$

Now we know that, 1 cm = 10000000 nm,

Converting the wavelength from Cm → Nm

$\lambda \: = 1.282 \times {10}^{ - 4} \times {10}^{7}$

$\lambda \: = 1.282 \times {10}^{ - 4 + 7}$

$\sf \lambda \: = 1.282 \times {10}^{ 3} nm$

$\sf \small Thanks \: for \: joining \: brainly \: community!$

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