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serg [7]
3 years ago
6

Increase the brightness of the light for the colors red, yellow, orange, and green. What effect did the brightness of the light

have on the metal?
(science)
Chemistry
2 answers:
vichka [17]3 years ago
7 0

Answer:

umm i really do not know sorry

NNADVOKAT [17]3 years ago
5 0
Hmmm.......... .......
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Which statement is true about the elements in the periodic table?
aleksandr82 [10.1K]
The correct answer is elements in family 7 have similar properties because elements in the same group or family have the same number of valence electrons.
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Which of the answer choices shows the correct Lewis dot diagram for hydrogen cyanide?
masya89 [10]
The Lewis dot diagram structure for hydrogen cyanide would be the following:

H - C --- N

--- = triple bond.
6 0
4 years ago
What is happening in the diagram? ​
OlgaM077 [116]

Answer:It is a picture with a boat sailing on the water with sound waves going down to an old and sunken ship but the waves then go back up to the boat in a different angle.

Explanation:

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4 0
4 years ago
When a metal was exposed to light at a frequency of 4.07× 1015 s–1, electrons were emitted with a kinetic energy of 3.30× 10–19
Licemer1 [7]

Answer :  The maximum number of electrons released = 1.432\times 10^{12}electrons

Explanation : Given,

Frequency = 4.07\times 10^{15}s^{-1}

Kinetic energy = 3.30\times 10^{-19}J

Total energy = 3.39\times 10^{-7}J

First we have to calculate the work function of the metal.

Formula used :

K.E=h\nu -w

where,

K.E = kinetic energy

h = Planck's constant = 6.626\times 10^{-34}J/s

\nu = frequency

w = work function

Now put all the given values in this formula, we get the work function of the metal.

3.30\times 10^{-19}J=(6.626\times 10^{-34}J/s\times 4.07\times 10^{15}s^{-1})-w

By rearranging the terms, we get

w=2.367\times 10^{-18}J

Therefore, the works function of the metal is, 2.367\times 10^{-18}J

Now we have to calculate the maximum number of electrons released.

The maximum number of electrons released = \frac{\text{ Total energy}}{\text{ work function}}

The maximum number of electrons released = \frac{3.39\times 10^{-7}J}{2.367\times 10^{-19}J}=1.432\times 10^{12}electrons

Therefore, the maximum number of electrons released is 1.432\times 10^{12}electrons

8 0
3 years ago
Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K sp = 1.4 × 10 –5). 1.5 × 10–2 M 1.4 × 10–
spin [16.1K]

Answer:

3.0x10⁻²M

Explanation:

Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:

Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻

When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:

ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:

1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

1.4x10⁻⁵ = [2X]² [X]

1.4x10⁻⁵ = 4X³

3.5x10⁻⁶ = X³

0.015 = X

As [Ag⁺] is 2X:

[Ag⁺] = 0.030 = 3.0x10⁻²M

The answer is:

<h3>3.0x10⁻²M</h3>
7 0
3 years ago
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