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nordsb [41]
3 years ago
5

9.

Chemistry
1 answer:
777dan777 [17]3 years ago
7 0

Answer:

salt in the Dead Sea is sodium chloride

while salt in the lab is either soluble or

insoluble in water.

Explanation:

The term 'salt' in the school laboratory does not always refer to sodium chloride. It is a generic term used for many substances especially those substances formed by neutralization reaction.

There are many salts that are used in the laboratory. Some of these salts are soluble in water while some are not soluble in water.

Salt in the dead sea always refers to sodium chloride, hence, salt in the dead sea is different from salt in the school laboratory.

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All chemical reactions have a conservation of(1) mass, only(2) mass and charge, only(3) charge and energy, only(4) mass, charge,
Semenov [28]
All chemical reactions have a conversation of mass and energy. 

Because:- There are only two laws for conversation in a chemical reaction. The conversation of mass, no mass can be created nor destroyed. Also, the law of conversation of mass states that no energy can be created nor destroyed in a chemical reaction. The charge can obviously change because they can bond and change charges. 
7 0
3 years ago
Read 2 more answers
Calculate the density of sulfuric acid if 35.4 mL of the acid weighs 65.14 g
Nadya [2.5K]
M/V=D
65.14/35.4≈1.84
The density of the sulfuric acid would be about 1.84g/mL
5 0
3 years ago
When sodium is excited in a flame, two ultraviolet spectral lines at (wave length symbol)=372.1 nm and (wave length symbol)=376.
uranmaximum [27]
<span>Higher energy = shorter wavelength
Frequency is one cycle over an amount of time (seconds)
So higher frequency = higher energy = shorter wavelength</span>
5 0
3 years ago
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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
Why does nitric oxide (no) act as a paracrine signal that affects only neighboring cells?
VMariaS [17]

The correct answer is due to rapid conversion of nitrates into nirites in extracellular fluids.

Due to the fact that it is quickly transformed to nitrates and nitrites in the extracellular fluid, nitric oxide (NO) functions as a paracrine signal that only impacts nearby cells. Because it relaxes the smooth muscle cells in blood vessel walls, nitric oxide (NO) causes blood vessels to widen. Cell signaling is a type of cellular communication in which a cell produces a signal to cause changes in neighboring cells, changing the behavior of those cells. Paracrine signaling is one type of cell signaling. Responses to allergens, tissue repair, the development of scar tissue, and blood clotting are a few examples of paracrine signaling. The transmission of signals through synapses between nerve cells is known as paracrine signaling.

Learn more about paracrine signal here :-

brainly.com/question/12538424

#SPJ4

5 0
1 year ago
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