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3 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]3 years ago

6 0

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

Setler [38]3 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 $Mg^{2+}\text{ and }Br^-$

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elena-s [515]

Answer:

Explanation:

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2 years ago

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A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several

mamaluj [8]

Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

$E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

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3 years ago

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction

steposvetlana [31]

Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

Where:

vavg = average velocity.

Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

Have a nice day!

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3 years ago

A beam of protons enter the electric field of magnitude E = 0.5 N/C between a pair of parallel plates. There is a magnetic field

HACTEHA [7]

Answer:

0.217 m/s

Explanation:

The protons in the beam passes undeflected when the electric force is equal to the magnetic force:

qE = qvB

where

q is the proton's charge

E is the magnitude of the electric field

v is the speed of the protons

B is the magnitude of the magnetic field

Re-arranging the equation,

$v=\frac{E}{B}$

And by substituting

E = 0.5 N/C

B = 2.3 T

We find

$v=\frac{0.5}{2.3}=0.217 m/s$

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3 years ago

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yKpoI14uk [10]

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