3. Beta decay is when a proton and ______ are emitted from a neutron.

Determine an empirical formula from each set of

Lera25 [3.4K]

Answer:

59.30% I

Explanation:

mi respuesta es esa porque es la más sercana

5 0

3 years ago

Draw the product formed when cyclohexene is reacted with H2 in the presence of Pt. Note: If adding hydrogen atoms to a carbon at

tigry1 [53]

Answer:

It has been drawn and uploaded as an attachment. Please download it to see the structure.

Explanation:

The product formed as a result of the reaction of cyclohexene with H2 in presence of Pt (platinum) can be described as catalytic hydrogenation. Catalytic hydrogenation is defined as the process of hydrogen addition in the presence of a catalyst, which in this case is platinum.

Note that Cyclohexene (alkene) is a hydrocarbon molecule represented by the chemical formula, C6H10 .

It consists of a double bond. During the hydrogenation reaction, the alkene undergoes an addition reaction to give alkane which is a saturated hydrocarbon as the product.

The first step in order to derive the product is to draw the chemical structure of cyclohexene and identify the double bond present in it.

The final product can be derived by replacing the double bond with the single bond and satisfying all the valences of the carbon atom. The final product structure has been drawn and uploaded as an attachment. Please download it to see the structure.

Ans:

The structure of the cyclohexane thus, formed has been shown as follows with all the hydrogen atoms:

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

3 years ago

The article explains that elephants sometimes communicate in low ultrasound. Based on what you learned above, would infrasound w

Likurg_2 [28]

I have not read the article, and so I don't know in what context this is referring to. But I do know that <span>Infra-sound is made up of a really low frequency sound, beyond the range of hearing by humans. </span>Helpful? :)

6 0

4 years ago

What is the total ionic equation for the following reaction?

schepotkina [342]

Answer: Option (d) is the correct answer.

Explanation:

An equation in which electrolytes are represented in the form of ions is known as an ionic equation.

Strong electrolytes easily dissociate into their corresponding ions. Hence, they form ionic equation.

$H_{2}CrO_{4}$ is a strong acid and $Ba(OH)_{2}$ is a strong bases, therefore, both of them will dissociate into ions.

Thus, total ionic equation will be as follows.

$2H^{+} + CrO_{4}^{-} + Ba^{2+} + 2OH^{-} \rightarrow Ba^{2+} + CrO_{4}^{-} + 2H_{2}O$

5 0

3 years ago

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