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3 years ago

Elaborate on the reason(s) that matter is said to move even as in a solid state.

1 answer:

3 0

1) The correct answer is
C) The particles are not able to move out of their positions relative to one another, but do have small vibrational movements.
In solids, in fact, particles are bound together so they cannot move freely. However, they can move around their fixed position with small vibrational movements, whose intensity depends on the temperature of the substance (the higher the temperature, the more intense the vibrations). For this reason, we say that matter moves also in solid state.

2) The correct answer is
A) increase the concentration of both solutions
In fact, when we increase the concentration of both solutions, we increase the number of particles that react in both solutions; as a result, the speed of the reaction will increase.

3) The correct answer is
C) gas → liquid → solid
In gases, in fact, particles are basically free to move, so the intermolecular forces of attraction are almost negligible. In liquids, particles are still able to move, however the intermolecular forces of attraction are stronger than in gases. Finally, in solids, particles are bound together, so they are not free to move and the intermolecular forces of attraction are very strong.

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Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m

AleksandrR [38]

Answer:

Part a)

$10\hat i + 15\hat j = \vec v$

Part b)

$\Delta K = 500 J$

Explanation:

As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved

so we can say

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v$

$5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v$

$10\hat i + 15\hat j = \vec v$

Part b)

magnitude of the initial speed of A = $\sqrt{15^2 + 30^2} = 33.54 m/s$

magnitude of the initial speed of B = $\sqrt{10^2 + 5^2} = 11.18 m/s$

magnitude of final speed of A = $\sqrt{5^2 + 20^2} = 20.61 m/s$

magnitude of final speed of B = $\sqrt{10^2 + 15^2} = 18.03 m/s$

Now change in total kinetic energy is given as

$\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)$

$\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)$

$\Delta K = 500 J$

6 0

3 years ago

What is the primary outgoing radiation put off by the sun?

ira [324]

I really don’t know but I think it’s D

7 0

3 years ago

1 point

Svetllana [295]

Answer:

Beta 17,000K, bc warmer is more blue

5 0

3 years ago

a distant galaxy is studied with a radio telescope x ray telescope and optical light telescope. the images form which set of mus

aivan3 [116]

Answer:

The modern instruments or we can say the different levels of telescopes are used to explore and study the distant galaxies. i.e the Hubble telescope is out there providing the data regarding the different properties of the celestial entities which in other case is not visible to the human naked eye.

Explanation:

Scientists and research workers are in constant search for more answers as they explore the universe and implement the laws of physics on the celestial entities. But, most of the objects inside the universe are not visible to human naked eye, as they are far from sight and thus more advanced form of instruments like the x-ray, optical, and light telescopes are used to determine the different properties of the celestial entities inside the universe.

As, these telescopes includes the most recent "Hubble telescope", which is out there inside the space to explore the universe and more over the galaxies by subjecting them with x-rays and then provide us with a very rough but valid results to study the distant galaxies.

6 0

3 years ago

What molecular geometry would be expected for f2 and hf?

Kaylis [27]

The molecular geometry of both F2 and HF is linear.

There are only two atoms which are covalently bonded and thus, the bonding scheme with the atoms looks like this;

F --- F

H---F

So, both are linear.

4 0

3 years ago