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sukhopar [10]
3 years ago
5

The study of the structure, reaction, and composition of substances is blank​

Chemistry
1 answer:
horrorfan [7]3 years ago
4 0
The answer is Chemistry.
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the term diffusion comes from a latin word meaning to spread apart. how is the term diffusion related to its latin word of origi
Nitella [24]

it means basically the same thing but can be used in different context. like if someone says, "lets diffuse the situation." it means lets separate it and find out what's going on with it. Like if you don't understand a sentence when you were younger your teacher would tell you to break down the sentence, and try to use context clues. What you are doing is diffusing the sentence so that your not looking at it as one big confusing sentence, your "breaking it down" and that helps you a lot. I hope I helped you, if not let me know and I can diffuse the topic a little more to make it more approachable and easy to understand. :)

3 0
3 years ago
Read 2 more answers
A compound has 15.39 g of gold for every 2.77 g of chlorine.
Ugo [173]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The question is to determine the simplest mass ratio of gold to Chlorine in the compound.

Since the mass of gold in the compound compared to chlorine is 15.39 g for every 2.77 g, the mass of gold per gram of chlorine is given as:

15.39 / 2.77 = 5.56 g of gold to two decimal places

<em>Therefore, for every 5.56 g of gold, there is 1 g of chlorine.</em>

<em>Note : The ratio in which different elements combine by mass to form a compound is given by the law of constant composition which states that, "all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass.</em>

3 0
3 years ago
Which of the following statements is not correct?
Lilit [14]

Answer:

It's 4.1 moles Li2O = 94 g on Odyssey ware

8 0
3 years ago
Read 2 more answers
A chemistry student adds a quantity of an unknown solid compound X to 5.00 L of distilled water at 15.° C . After 10 minutes of
Yuliya22 [10]

Answer:

34 g/100 mL

Explanation:

The solubility of a compound can be expressed in g/100mL, for this we must divide the mass of the compound that dissolves in the solute by the volume of the solvent.

The solvent, in this case, is water, and that mass of the solute X that dissolved is the mass that was recovered after the solvent was drained and evaporated. So the solubility of X (S) is:

S = 0.17 kg/5L

S = 170g/5000mL

S = 170g/(5*1000)mL

S = 34 g/100 mL

8 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
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