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DedPeter [7]
3 years ago
15

In each row check off the boxes that apply to the highlighted reactant. reaction The highlighted reactant acts as a... (check al

l that apply)
1. HCH3CO2(aq) + NH3(aq) → CH3CO−2(aq) + NH+4(aq)
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base

2. BH3(aq) + NH3(aq) → BH3NH3(aq)
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base

3. HNO2(aq) + C2H5NH2(aq) → NO−2(aq) + C2H5NH+3(aq)
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base
Chemistry
1 answer:
tekilochka [14]3 years ago
7 0

The given question is incomplete. The complete question is :

In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)

1. HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3COO^-(aq)+NH_4^+(aq)

here underlined is HCH_3CO_2

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

2. BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)

Here underlined is NH_3

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

3. HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)

Here underlined is C_2H_5NH_2

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

Answer: 1. Brønsted-Lowry acid

2. Lewis base

3. Brønsted-Lowry 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.

1.  HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3CO^{2-}(aq)+NH_4^+aq)

As HCH_3CO_2(aq) is donating a proton , it acts as a bronsted acid.

2. BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)

As NH_3 contains a lone pair of electron on nitrogen , it can easily donate electrons to BH_3 and act as lewi base.

3.  HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)

As C_2H_5NH_2(aq) is accepting a proton , it acts as a bronsted base.

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How many atoms are in 165 g of calcium?
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Atomic mass Calcium = 40.078 a.m.u

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2 years ago
When oxygen is available what happens immediately after glycolysis?
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Answer:

NADH is formed

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3 0
3 years ago
The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe
horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

-  (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

-  (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

=  0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

-  (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

-  (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

-  (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0
2 years ago
I have a question about ionization energy.
ValentinkaMS [17]
Ionization energy is the measure of the extend to which the nucleus attracts the outermost electron
if ionization energy us high than force of attraction Is high so it is not easy to remove and vice versa .
hope you understand.....
5 0
3 years ago
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