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Genrish500 [490]

3 years ago

12

An oxygen atom picks up two additional, free-floating electrons. Is the charge of the newly formed oxygen ion positive, negative

, or neutral?

1 answer:

MariettaO [177]3 years ago

4 0

Answer: Negative

Explanation:

An oxygen atom has the chemical symbol O. Its atomic number is 8, with an electronic configuration of 1s2, 2s2 2p4.

Now, due to the incompletely filled p-orbital, oxygen picks up two additional, free-floating electrons (2e-) to complete its outermost shell. And since each electrons received carries a negative charge, oxygen atom becomes a negatively charged divalent ion, O2-

See the equation below

O + 2e- ---> O2-

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The period of a sound wave is 0.002 seconds. The speed of a sound is 344 m/s. Find the frequency and wavelength of the sound wav

ankoles [38]

Answer:

I think that's the solution

5 0

3 years ago

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

3 years ago

A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic

Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light $c= 2.99793\times10^{8}\ m/s$

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

$\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}$

Put the value into the formula

$\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}$

$\Delta l=1\sqrt{1-(0.96)^2}$

$\Delta l=0.28\ m$

We need to calculate the time the stick take to move

Using formula of time

$t=\dfrac{\Delta l}{v}$

Put the value into the formula

$t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}$

$t=9.72\times10^{-10}\ sec$

$t=0.97\ ns$

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0

3 years ago

Im back and need help

const2013 [10]

Always here to help. Bring it!!!

7 0

3 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ???? 0 ν0 . Find the mi

Allisa [31]

Answer:

<h2>E = 2.8028*10⁻¹⁹ Joules</h2>

Explanation:

The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo

h = planck's constant

fo = threshold frequency

Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹

h = 6.626× 10⁻³⁴ m² kg / s

Substituting this value into the formula to get the energy E

E = 4.23×10¹⁴ * 6.626 × 10⁻³⁴

E = 28.028*10¹⁴⁻³⁴

E = 28.028*10⁻²⁰

E = 2.8028*10⁻¹⁹ Joules

7 0

4 years ago