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Vinil7 [7]
3 years ago
14

HELPPPPP PLEASEEE ILL MARK YOU BRAINLIESTTTTTT!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry
2 answers:
azamat3 years ago
6 0

Answer:

Ionic bonding occurs when atoms either gain or lose one or more valence electrons, resulting in the atom having either a negative or positive charge.

Through ionic bonding, an atom of each element will combine with the other to form a molecule, which is more stable since it now has a zero charge.

Explanation:

soldier1979 [14.2K]3 years ago
3 0

Answer:

Ionic bonding occurs when atoms either gain or lose one or more valence electrons, resulting in the atom having either a negative or positive charge.

Explanation:

Hope this helps!!:)

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How can forgery be used in crime
Tems11 [23]

Some bad people will tend to forge dollar bills and make fake bills, giving them to cashiers and asking for it to be split into a different amount of bills so they are given real cash and wont be caught or they will buy things, return them and be given back real money and get away with it. This is obviously a crime because it's fake.

3 0
3 years ago
Read 2 more answers
Consider the incorrectly balanced combustion equation:
Ulleksa [173]
Oxygen is balanced incorrectly.
The correct balanced equation: 2C6H6+15O2=12CO2+6H2O
6 0
3 years ago
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0
3 years ago
The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O represents an acid-base tritration. The equivalence point occurs when 24.75 mL o
artcher [175]
            moles NaOH = c · V = 0.2432 mmol/mL · 24.75 mL = 6.0192 mmol
            moles H2SO4 = 6.0192 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.0096 mmol
Hence
            [H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
The answer to this question is  [H2SO4] = 0.07729 M

6 0
3 years ago
In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
crimeas [40]

Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

5 0
4 years ago
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