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Alex Ar [27]
3 years ago
14

What is the mass of oxygen in 10g of water

Chemistry
1 answer:
Doss [256]3 years ago
8 0

Answer:

8.90g

Explanation:

2×O=2×16.00=32.00g

mass=0.278×32.00g=8.90g

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From the list below, choose which groups are part of the periodic table. metals acids flammable gases nonmetals semimetals ores
iris [78.8K]

The periodic table of the elements are describe the electronic configuration of the elements on which the properties of the elements depends. Among the given groups only metal, non-metal and semi-metal group are the part of periodic table. The metallic property depends upon the binding energy of the electrons with the nucleus. Thus the elements which have the valence electrons more near to the nucleus that is s-block elements are more metallic in nature. On the other hand the elements which have the valence electrons far from the nucleus are more non-metallic in nature like p-block elements. However the binding energy or the attraction of the outermost electrons to the nucleus depends not only its valence electrons position but also some other factors like shielding effect, effective nuclear charge etc.

The elements which are in between the metals and non-metals can be classified as semi-metals.

Although the conductivity of a material is an inherent property of the metals but sometime the nonmetals or semi-metals are also behave like a conductor due to presence of the other elements, thus it cannot be a p[property of the periodic table. Similarly acidity, flammable gases are not part of the periodic table.      

8 0
3 years ago
Read 2 more answers
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
What is the chemical test for chlorine?
Ann [662]

Answer:

hope it helps you a little

3 0
3 years ago
How much does the air weigh within a column that is 1 square inch in area that extends from sea level all the way to the top of
fredd [130]

Answer:

14.7 lbs

Explanation:

Air pressure is the weight of the air above us. It is approximately 14.7 pounds or lbs per square inch at sea level. It means that an air column weights 14.7 lbs, 1 square inch in diameter, reaching all the way up to the top of the atmosphere.

5 0
3 years ago
Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?
lina2011 [118]

Answer: k_a for C_6H_5NH_3^+ at 25°C is 0.588\times 10^{-4}

Explanation: We are given k_b of C_6H_5NH_2 at 25°C which is 1.7\times 10^{-9}

To calculate the k_a of C_6H_5NH_3^+, we use the formula:

k_w=k_a\times k_b

k_w\text{ at }25^o=1\times 10^{-14}

Putting values in above equation, we get:

1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}

7 0
3 years ago
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