All categories

2 years ago

Which pair of ions is most likely to form an ionic bond? A. H- and H- B. Mg2+ and Br- C. N3- and O2- D. Na+ and Al3+

Physics

2 answers:

Georgia [21]2 years ago

6 0

$\textsf{answer is b .}\\\textsf{because magnesium is very hungry}\\\textsf{so mg takes 2electrons}$

Setler [38]2 years ago

3 0

Answer: The pair of ions that is most likely to form an ionic bond is $Mg^{2+}\text{ and }Br^-$

Explanation:

Ionic compound is defined as the compound which is formed by complete transfer of electrons from one atom to another atom.

The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal (carrying positive charge) and a non-metal (carrying negative charge)

Ionic compound is formed by the attraction of oppositely charged ions.

For the given options:

Magnesium ion $(Mg^{2+})$ and bromine ion $(Br^-)$ will attract each other because they are oppositely charged and will lead to form ionic compound.

The ionic compound formed is magnesium bromide $(MgBr_2)$

Hence, the pair of ions that is most likely to form an ionic bond is $Mg^{2+}\text{ and }Br^-$

You might be interested in

When heat flows from one substance to another, what happens to the temperature of the substance giving off the heat and to the t

blagie [28]

The temperature of the substance giving off the heat decreases while the temperature of the substance receilving the heat increases. they leach what is called equlibrium point where heat energy can longer be exchanged hence equql temperature. this isThermal physics

8 0

3 years ago

Match the letters to the numbers for 30 points!

liq [111]

-- 'Ca' (Calcium) is an element.
-- The proton has a positive charge.
-- Nuclear fusion results in the synthesis of atoms of new elements.
-- H₂O (water) is a chemical compound.
-- Nuclear fission is a decay of the nucleus.
-- The atomic number of an element is the number of protons
in each atom of it.
-- I suppose you're using the Greek letter η ('eta', not 'nu')
to represent the neutron.
-- I suppose you're using ' e ' to represent the electron.

7 0

3 years ago

Read 2 more answers

A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean, when its brakes fai

patriot [66]

Answer:

a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s

b) The time it takes the car to reach the edge is 4.79 s

c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal

d) The total time interval the car is in motion is 6.34 s

e) The car lands 24 m from the base of the cliff.

Explanation:

Please, see the figure for a description of the situation.

a) The equation for the position of an accelerated object moving in a straight line is as follows:

x =x0 + v0 * t + 1/2 a * t²

where:

x = position of the car at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:

x = 1/2 a * t²

With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

t = 4.79 s

The equation for velocity is as follows:

v = v0 + a* t

Where:

v = velocity

v0 = initial velocity

a = acceleration

t = time

For the car, the velocity will be

v = a * t

at the edge, the velocity will be:

v = 4.05 m/s² * 4.79 s = 19.4 m/s

b) The time interval was calculated above, using the equation of the position:

x = 1/2 a * t²

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

t = 4.79 s

c) When the car falls, the position and velocity of the car are given by the following vectors:

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

v =(v0x, v0y + g * t)

Where:

r = position vector

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity

v = velocity vector

First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.

Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:

cos -37.0º = v0x / v0

(the angle is negative because it was measured clockwise and is below the horizontal)

(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)

v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s

sin 37.0º = v0y/v0

v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s

Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):

y = y0 + v0y * t + 1/2 * g * t²

Notice in the figure that the y-component of the vector "r final" is -30 m, then:

-30 m = y0 + v0y * t + 1/2 * g * t²

According to our reference system, y0 = 0:

-30 m = v0y * t + 1/2 g * t²

-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²

0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²

Solving this quadratic equation:

t = 1.55 s ( the other value was discarded because it was negative).

Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:

vy = v0y + g * t

vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s

The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).

The velocity vector when the car lands is:

v = (15.5 m/s, -26.9 m/s)

We have to express it in magnitude and direction, so let´s find the magnitude:

$|v| = \sqrt{(15.5 m/s)^{2} + (-26.9 m/s)^{2}} = 31.0m/s$

To find the direction, let´s use trigonometry again:

sin α = vy / v

sin α = 26.9 m/s / 31.0 m/s

α = 60.2º

(notice that the angle is measured below the horizontal, then it has to be negative).

Then, the vector velocity expressed in terms of its magnitude and direction is:

vy = v * sin -60.2º

vx = v * cos -60.2º

v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)

The velocity is 31.0 m/s at 60.2º below the horizontal

d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.

total time = falling time + rolling time

total time = 1,55 s + 4.79 s = 6.34 s

e) Using the equation for the position vector, we have to find "r final 1" (see figure):

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

Notice that the y-component is 0 ( figure)

we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.

r final 1 = ( v0x * t, 0)

r final 1 = (15.5 m/s * 1.55 s, 0)

r final 1 = (24.0 m, 0)

The car lands 24 m from the base of the cliff.

PHEW!, it was a very complete problem :)

5 0

2 years ago

A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a

scZoUnD [109]

Hi there!

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
$T_h = \frac{T}{2}$