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3 years ago

Which elements are used for artistic objects

Chemistry

2 answers:

JulijaS [17]3 years ago

7 0

Color, Line, Texture, Space, form, and value

andrew11 [14]3 years ago

6 0

The elements used for artistic objects are components like color, form, line, shape, space, texture, and value.

Explanation:

They forms the basic building blocks of art. A two-dimensional painting or three-dimensional sculpture are decided by the artist. Line, shape, form may encloses volume a such as cube, cylinder, sphere, pyramid.

Value defines the shades of the colors. Spaces are defined the depthness. Colour made up hue having the name of colour value being the darkness or brightness of the colour. Intensity describes the quality of purity and brightness .Texture describes how things appear or feel touched.

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The photon of light that is emitted as an electron drops back to its original orbit is:

Alex73 [517]

Answer:

energy

Explanation:

The photon of light that is emitted as an electron drops back to its original orbit is energy and this energy is released during de-excitation process.

The electron is jumped into higher level and back into lower level by absorbing and releasing the energy.

The process is called excitation and de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits. For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy because electron can not stay longer in higher energy level and comes to ground state.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum

3 0

3 years ago

At 850K, 65L of gas has a pressure of 450kPa. What is the volume (in liters) if the gas is cooled to 430K and the pressure decre

Novosadov [1.4K]

Answer:

V₂ = 45.53 L

Explanation:

Given data:

Initial temperature = 850 K

Initial volume = 65 L

Initial pressure = 450 KPa

Final temperature = 430 K

Final pressure = 325 KPa

Final volume = ?

Solution:

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 450 KPa× 65 L × 430 K / 850 K × 325KPa

V₂ = 12577500 KPa .L. K / 276250 K. KPa

V₂ = 45.53 L

8 0

3 years ago

An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ

MAVERICK [17]

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = 0,1122 g of C ≡ 112,2mg of C

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = 0,0110 g of H ≡ 11,0mg of H

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = $\frac{112,2 mgC}{200,0mg}$ ×100 = 56,1%C

%H = $\frac{11,0 mgH}{200,0mg}$ ×100 = 5,5%H

%Cl = $\frac{55,14 mgCl}{200,0mg}$ ×100 = 27,6%Cl

%N = $\frac{21,66 mgN}{200,0mg}$ ×100 = 10,8%N

I hope it helps!

3 0

3 years ago

2. (Exercise 4.1.6) A liquid adhesive consists of a polymer dissolved in a solvent. The amount of polymer in the solution is imp

Lilit [14]

Answer:

800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .

Explanation:

3000 lb of 13% solution is required .

Total adhesive in weight = 3000 x .13 = 390 lb of adhesive

Available = 500 lb of 10% solution = 50 lb of adhesive

Rest = 390 - 50 = 340 lb required .

rest mass of solution = 3000 - 500 = 2500 lb

mass of adhesive required = 340 lb

Let the mass of 20% required be V

mass of adhesive = .20 V

.20 V = 340

V = 1700

rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent

So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .

6 0

4 years ago

Two oceanic plates at a transform boundary loctaion 6

matrenka [14]

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4 0

3 years ago