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Makovka662 [10]
3 years ago
10

What conclusion can be drawn from the statement that an element has high electron affinity, high electronegativity, and a high i

onization energy?
A) The element is most likely from Group 1A or 2A and in period 1 or 2.

B) The element is most likely from Group 2A or 3A and in period 6 or 7.

C) The element is most likely from Group 4A or 5A and in period 4 or 5.

D) The element is most likely from Group 6A or 7A and in period 2 or 3.
Physics
1 answer:
amid [387]3 years ago
4 0

Answer:

D) The element is most likely from Group 6A or 7A and in period 2 or 3.

Explanation:

Electronegativity of an atom is the tendency of an atom to attract shared paired of electron to itself. Electronegativity increase across the period from left to right.The ability of an atom to attract electron to itself is electronegativity. Group 7A and 6A elements can easily attract atoms to itself so they are highly electronegative. The most electronegative element in the periodic table is fluorine.Group 6A and 7A is likely to have high electronegativity.

Electron affinity of an atom is the amount of energy release when an atom gains electron . Generally, when atom gains electron they become negatively charged. Group 6A and 7A elements have high electron affinity.  

Ionization energy is the energy required to remove one or more electron from a neutral atom to form cations.  ionization energy of group 7A and 6A are usually high because the energy required to remove these electron is usually very high . The elements in this groups usually gain electron easily so the energy to remove electron is very high.

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If you start skating down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill, your
baherus [9]
Your potential energy at the top of the hill was (mass) x (gravity) x (height) .

Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .

If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.

(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)

Divide each side by 'mass' :

(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.

Multiply each side by 2 :

(speed)² = 2 x (gravity) x (height)

Take the square root of each side:

<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>

We could go one step further, since we know the acceleration of gravity on Earth:

Speed at the bottom = 4.43 x square root of (height of the hill)

This is interesting, because it says that a hill twice as high won't give you
twice the speed at the bottom.  The final speed is only proportional to the
<em>square root </em>of the height, so in order to double your speed, you need to
find a hill that's <em>4 times</em> as high.






6 0
3 years ago
during a crash a dummy with the mass of 60.0 kg hits a airbag that exerts a constant force in the dummy the acceleration of the
Yakvenalex [24]

Explanation:

F = ma

F = (60.0 kg) (250 m/s²)

F = 15,000 N

6 0
3 years ago
A cyclist intends to cycle up a 7.70o hill whose vertical height is 126m. Assuming the mass of bicycle plus person is 75.0kg, ca
Ksivusya [100]

Answer:

92704.5 J

596.44737 N

Explanation:

m = Mass of person + bicycle = 75 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Vertical height = 126 m

\theta = Angle = 7.7°

d = Diameter = 0.388 m

Work done against gravity is given by

P=mgh\\\Rightarrow P=75\times 9.81\times 126\\\Rightarrow P=92704.5\ J

Work done is 92704.5 J

Force required is given by

F=\dfrac{mgrsin\theta}{\pi d}\\\Rightarrow F=\dfrac{75\times 9.81\times sin7.7\times 5.12}{\pi\times 0.388}\\\Rightarrow F=596.44737\ N

The force is 596.44737 N

7 0
3 years ago
A box of books weighing 290 N is shoved across the floor of an apartment by a force of 400 N exerted downward at an angle of 34.
NARA [144]

Answer: 1.95s

Explanation:

Given

ma = 290 cos 34.9 - fk

fk = 290 cos 34.9 - ma

fn = mg + 400 sin Φ

fn = 290 + 400 sin 34.9

fn = 290 + 228.9

fn = 518.9

fk = fn * uk

uk = 0.57

290 cos 34.9 - ma = 518.9 * 0.57

290 cos 34.9 - ma = 295.8

290 cos 34.9 - 295.8 = ma

ma = -58

m = 290/10 = 29

a = 58/29

a = 2

Using equation of motion

S = ut + .5at²

3.8 = 0 + .5*2*t²

3.8 = t²

t = 1.95s

3 0
3 years ago
1.00kg of ice at -24 degrees Celsius is placed in contact with a 1.00kg block of a metal at 5.00 degrees Celsius. They come to e
Alona [7]

Answer:

C₂ = 2.22 KJ/kg °C

Explanation:

Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:

Heat\ Lost\ by\ Metal\ Block = Heat\ Gained\ by\ Ice\\m_{1}C_{1} \Delta T_{1} = m_{2}C_{2} \Delta T_{2}

where,

m₁ = mass of ice = 1 kg

C₁ = specific heat of ice = 2.04 KJ/kg.°C

ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C

m₂ = mass of metal block = 1 kg

C₂ = specific heat of metal = ?

ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C

Therefore, using these values in the equation, we get:

(1\ kg)(2.04\ KJ/kg.^0C)(15.12\ ^0C) = (1\ kg)C_{2}(13.88\ ^0C) \\C_{2} = \frac{30.84\ KJ}{13.88\ kg/^0C}

<u>C₂ = 2.22 KJ/kg °C</u>

4 0
3 years ago
Read 2 more answers
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