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Veseljchak [2.6K]
3 years ago
10

Mass of 2 x 1021 number of atoms of a

Physics
1 answer:
tangare [24]3 years ago
3 0

Answer: The mass of 0.5 mole of element ab is 81.67 g.

Explanation:

According to the mole concept, 1 mole of a substance contains 6.022 \times 10^{22}. As, the mass of 2 \times 10^{21} atoms is 0.49 g.

So, \frac{2 \times 10^{21}}{6.022 \times 10^{23}} mol = 0.49 g

0.003 mol = 0.49 g

Therefore, mass of 1 mole = \frac{0.49}{0.003}

                                            = 163.33 g/mol

Hence, mass of 0.5 mole of ab element is calculated as follows.

0.5 mol \times 163.33 g/mol

= 81.67 g

Therefore, the mass of 0.5 mole of element ab is 81.67 g.

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A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
Earth’s largest ecosystems, biomes, are defined primarily by:
Darina [25.2K]

Answer: e. All of the above.

Explanation:

Rainfall, temperature, seasonal variability are the important factors which determines the type of vegetation will grow in a biome, the type of animals will adapt and survive. Also these factors also determine the fact that type of landforms will form over a region. These factors are necessary for the development of the biodiversity. A biodiversity can be define as the variability and variety of life forms that can be found in the ecosystem or biome.      

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4 years ago
A certain atom has 26 protons, 26 electrons, and 30 neutrons. It mass number is?
pishuonlain [190]

The most common atom of iron has 26 protons and 30 neutrons in its nucleus. What are its atomic number, atomic mass, and number of electrons if it is electrically neutral? This atom has atomic number 26, atomic mass 56, and has 26 electrons.

4 0
3 years ago
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Using Newton's 2nd Law of Motion Formula (F=MA) answer the following.
vladimir1956 [14]

Explanation:

1200 is your answer for this question

6 0
3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
guajiro [1.7K]

Answer:

f_{e} = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

        M = M₀ m_{e}

        M = - L/f₀  25/f_{e}

Where f₀ and f_{e} are the focal lengths of the lens and eyepiece, respectively, all values ​​in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece (f_{e})

         f_{e} = - L / f₀ 25 / M

Let's calculate

        f_{e} = - 16 / 0.6 25 / (-400)

        f_{e} = 1.67 cm

The minus sign in the magnification is because the image is inverted.

          f_{e} = 1.7 cm

6 0
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