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cupoosta [38]
3 years ago
10

The study of the isolation, structure, properties, and uses of metals are called

Chemistry
2 answers:
torisob [31]3 years ago
7 0

Answer:

Metallurgy  is the correct answer.

Explanation:

Metallurgy is the branch of science which deal with the studies of physical and chemical properties of metals, its uses, and the production of metals.

It helps to know the physical and chemical behavior of metals.

Thus we can conclude that the study of the isolation, structure, properties, and uses of metals are called Metallurgy.

baherus [9]3 years ago
3 0

The answer is; metallurgy  

Metallurgy is a component of material science and engineering. This knowledge is also used to refine metals from their ores. The study of the physical nature of metals is called physical metallurgy while the study of the chemical nature of the metals is called chemical metallurgy.


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2.4 g of a solid is added to 91 g of water. What is the mass % concentration of the solution?
vladimir2022 [97]

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2.64%

Explanation:

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8 0
1 year ago
What happens to an atom if it loses an electron?
jolli1 [7]

Answer:

When an atom loses electron(s) it will lose some of its negative charge and so becomes positively charged. A positive ion is formed where an atom has more protons than electrons. In the opposite case when an atom gains electron(s) it becomes negatively charged (more electrons than protons).

Explanation:

7 0
3 years ago
Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

#SPJ4

8 0
2 years ago
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