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3 years ago

Explain the process of ionic bond formation between K (potassium, a metal) and Br (bromine, a nonmetal).

2 answers:

5 0

Potassium belongs to group IA of the elements. This means that it will give up one of its electrons to form the cation K+. Opposite to that is bromine in which it accepts one electrons to form the anion Br-. The binding of these elements will form KBr and is formed from transfer of electron from one element to the other. This is the mechanism of ionic bond formation.

ra1l [238]3 years ago

5 0

ANSWER ON ED: K donates, or transfers, one electron to bromine, which has 7 electrons. Both K and Br are now stable with 8 electrons. K becomes a positive ion and Br becomes a negative ion. The positive K ion and the negative Br ion attract each other to form an ionic bond.

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A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0

4 years ago

Which would hit the ground forest if dropped from the same height in a vaccum , a flat piece of paper or blowing ball?

yan [13]

Answer:

We conclude that if a flat piece of paper or blowing ball is dropped from the same height, They would hit the ground at the same time.

Thus,

option a) is correct.

Explanation:

When there is no air in the space, it is basically a vacuum.

In other words, the vacuum can be termed as a 'space devoid of matter'.

So, when there is no air in space, it means there is no force that can act on the body.

Therefore, no matter what objects (heavy or light) we may drop from the same height, they would all hit the ground at the same time.

As there is no force that can act on an object in the vacuum, it means every object in the vacuum seems to stand still while falling.

Therefore, we conclude that if a flat piece of paper or blowing ball is dropped from the same height, They would hit the ground at the same time.

Thus,

option a) is correct.

4 0

3 years ago

A hole of radius r has been drilled in a circular, flat plate of radius R. The center of the hole is a distance d from the cente

Setler [38]

Answer:

I_total = ½M[R² - ((r²/R²)(r² + 2d²))]

Explanation:

From the question, we have to consider the hole as an object of negative mass. Thus, the total inertia of the object will be;

I_total = I_plate - I_hole

Moment of inertia of a Disk is ½MR²

We are told it rotates around it's axis. Hence, we will use parallel axis theorem and we have;

I_plate = ½M_plate•R²

I_hole = ½M_hole•r² + M_hole•d²

I_hole = ½M_hole(r² + 2d²)

Thus;

I_total = ½M_plate•R² - ½M_hole(r² + 2d²)

Now,since the plate is uniform, then;

M_hole/M_plate = A_hole/M_plate

A_hole = πr²

A_plate = πR²

Thus;

M_hole/M_plate = πr²/πR²

M_hole/M_plate = r²/R²

M_hole = M_plate(r²/R²)

Let's put M_plate(r²/R²) for M_hole in the I_total equation to get;

I_total = ½M_plate•R² - (½M_plate(r²/R²)(r² + 2d²))

Factorizing, we have;

I_total = ½M_plate[R² - ((r²/R²)(r² + 2d²))]

We are told that mass of the solid disk before the hole is removed was M.

Thus;

I_total = ½M[R² - ((r²/R²)(r² + 2d²))]

4 0

3 years ago

A 0.750-kg object hanging from a vertical spring is observed to oscillate with a period of 1.50 s. When the 0.750-kg object is r

lawyer [7]

Answer:

New time period, $T_2=2.12\ s$

Explanation:

Given that,

Mass of the object 1, $m_1=0.75\ kg$

Time period, $T_1=1.5\ s$

If object 1 is replaced by object 2, $m_2=1.5\ kg$

Let $T_2$ is the new period of oscillation.

The time period of oscillation of mass 1 is given by :

$T_1=2\pi \sqrt{\dfrac{m_1}{k}}$

$1.5=2\pi \sqrt{\dfrac{0.75}{k}}$ ............(1)

The time period of oscillation of mass 2 is given by :

$T_2=2\pi \sqrt{\dfrac{m_2}{k}}$

$T_2=2\pi \sqrt{\dfrac{1.5}{k}}$ ............(2)

From equation (1) and (2) we get :

$(\dfrac{T_1}{T_2})^2=\dfrac{m_1}{m_2}$

$(\dfrac{1.5}{T_2})^2=\dfrac{0.75}{1.5}$

$\dfrac{1.5}{T_2}=0.707$

$T_2=2.12\ s$

So, the new period of oscillation is 2.12 seconds. Hence, this is the required solution.

3 0

3 years ago

When the rudder moves to the left the plane will move to the right

cluponka [151]

Answer:

yes because it bends air to make it move that way just like a bicycle when you steer it

Explanation:

5 0

3 years ago