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Tju [1.3M]
3 years ago
14

Scientists are trying to understand the composition of substances on

Chemistry
1 answer:
ahrayia [7]3 years ago
3 0
I believe the answer is b or c. good luck.
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Which of the following lists elements from lowest to highest atomic number
vitfil [10]
Lowest is Hydrogen highest is <span>Beryllium
-HOPE THIS HELPED </span>
8 0
4 years ago
The chemical formula for copper (II) phosphate is Cu3(PO4)2. What is the charge on each copper ion?
mrs_skeptik [129]
3+ because I just know it 8757890
6 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
200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentratio
Ede4ka [16]

Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

First we <u>calculate how many moles of each reagent were added</u>:

  • HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
  • Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:

  • 203.85 mmol HCl * \frac{1mmolBa(OH)_{2}}{2mmolHCl}= 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

  • 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

There are two OH⁻ moles per Ba(OH)₂ mol:

  • OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):

  • 716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

7 0
3 years ago
Draw lewis formula and structural formula of hydrogen ion please and thank you
tresset_1 [31]

The Lewis formula refers to a diagram showing the distribution of electrones and in case of a molecule it also shows the bonds.

The structural formula on the other hand is a representation of the molecular structure that shows all the atoms that form the molecule, arranged in a three dimentional space,

In this case we have the hydrogen ion, which is the simpliest case we can have.

Hydrogen ion is the hydrogen atom possitively charged as it has lost his electron. Therefore the structural formula is simply the following:

The Lewis formula is also very simple as this ion has no electrons and has no bonding to other atoms:

4 0
1 year ago
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