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4 years ago

Consider the following electron configurations to answer the question:

Physics

1 answer:

Anika [276]4 years ago

4 0

Answer:

(ii) 1s2 2s2 2p6 3s2

Explanation:

Electron Affinity is the energy change that occur when an atom gains an electron.

X₍₉₎ + e⁻ → X⁻ ΔE = Eea

ΔE is change in energy

Eea is electron affinity

Often, electron affinity has negative energy values. The more negative the electron affinity, the easier it is to add an electron to a particular atom. Electron affinity increases across the period in the periodic table. However, there are few exceptions:

1. The electron affinities of group 18 (8A) elements are greater than zero. This is because the atom has a filled valence shell, an addition of electron causes the electron to move to a higher energy shell.

2. The electron affinities of group 2 (2A) elements are more positive because addition of an electron requires it to reside in the previously unoccupied p sub-shell.

3. The electron affinities of group 15 (5A) elements are more positive because addition of an electron requires it to be put in an already occupied orbital.

Applying these consideration to the elements given in the question:

(i) The sum of the electron in 1s²2s² 2p⁶ 3s¹ = 2+2+6+1 = 11 Sodium (Na)

(ii) The sum of the electron in 1s² 2s² 2p⁶ 3s² = 2+2+6+2 = 12 Magnesium (Mg)

(iii) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p¹ = 2+2+6+2+1 = 13 Aluminium (Al)

(iv) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁴ = 2+2+6+2+4 = 16 Sulfur (S)

(v) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁵ = 2+2+6+2+5 = 17 Chlorine (Cl)

The atom that is expected to have a positive electron affinity is Magnesium which is a group 2A element with electron configuration of 1s² 2s² 2p⁶ 3s².

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At a pressure 43 kPa, the gas in a cylinder has a volume of 12 liters. Assuming temperature remains the same, if the volume of t

Allisa [31]

P = 43 * 10^3
V = 12 L

P' =?
V' = 8 L

PV = P'V'
43*10^3* 12 = P' * 8

I know you can do it from here, mate ;)

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3 years ago

Is it possible that a 60kg boy riding a BMX bike could exert more pressure on a road surface than a 2500kg truck? Explain why.

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4 years ago

Sasha did an experiment to study the solubility of two substances. She poured 100 mL of water at 20 °C into each of two beakers

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Answer:

b is the higher solubility then A

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3 years ago

To understand how to find the velocities of objects after a collision.

trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

Part A: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

Part B: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$

$v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m.v}{m+m}$

v_1 = v_2 = $\frac{v}{2}$

Part C: Object 1 is 2m, object 2 is m and elastic collision:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = $\frac{2m - m}{2m + m } . v$

v_1 = $\frac{v}{3}$

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.2m}{2m+m}.v$

v_2 = $\frac{4v}{3}$

Part D: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = $v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m}{m+3m}.v$

v_1 = v_2 = $\frac{v}{4}$

5 0

4 years ago

Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.

SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
For the first collission, only mass 1 is moving before it, so we can write the following equation:

$p_{i} = p_{f} = m*v_{o} (1)$

Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

$p_{f1} = 2*m*v_{1} (2)$

From (1) and (2) we get:
v₁ = v₀/2 (3)
Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

$p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)$

Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

$p_{2} = 3*m*v_{2} (5)$

From (4) and (5) we get:
v₂ = v₀/3 (6)

Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

$p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)$

Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

$p_{3} = 4*m*v_{3} (8)$

From (7) and (8) we get:
v₃ = v₀/4
This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.

5 0

3 years ago