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cupoosta [38]
3 years ago
7

How many valence electrons are found in the outermost energy level of fluorine ( f.?

Chemistry
1 answer:
Paraphin [41]3 years ago
3 0
There are 7 valence electrons
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Given that the wavelength of a wave is 4.57 x 10^9 m, what must be the
r-ruslan [8.4K]

Answer:

Option d is correct option = Frequency of light is 6.56×10⁻² Hz

Explanation:

Given data:

Frequency of light = ?

wavelength of light = 4.57×10⁹m

Solution:

Formula:

Speed of light = wavelength × frequency

c = λ × f

f = c/λ

This formula shows that both are inversely related to each other.

The speed of light is 3×10⁸ m/s

Frequency is taken in Hz.

It is the number of oscillations, wave of light make in one second.

Wavelength is designated as "λ" and it is the measured in meter. It is the distance between the two crust of two trough.

Now we will put the values in formula.

f = 3×10⁸ m/s / 4.57×10⁹m

f = 0.656×10⁻¹s⁻¹

s⁻¹  = Hz

f = 0.656×10⁻¹  Hz or 6.56×10⁻² Hz

6 0
4 years ago
If a drop of blood is 0.05 mL, how many drops of blood are in a blood collection tube that holds 8 mL ?
inn [45]
160 drops of blood is needed 
6 0
3 years ago
Read 2 more answers
35 points for one question
True [87]

The result of the procedure is that the copper strip should go through oxidation.

<h3>How to explain the procedure?</h3>

The copper strip should be considered as the anode and the nail should be considered as the cathode.

The anode is the electrode of an electrochemical cell at which oxidation occurs. The anode should be + while the cathode should be negative. Both should be placed in the tank that should be filled with the electrolyte solution.

Learn more about oxidation on:

brainly.com/question/25886015

#SPJ1

4 0
2 years ago
Now molecules: ____ molecules of H2 + ____ molecules of O2 → ____molecules of H2O
Valentin [98]

Answer:

4 H2 + 1 O2 = 2 H2O

6 0
4 years ago
The Ksp for calcium fluoride, CaF2, is 1.5 x 10-10. a) Excess solid calcium fluoride is added to 1.00 L of pure water. Calculate
solniwko [45]

Answer:

a) [ Ca2+ ] = 3.347 E-4 mol/L

b) [ Ca2+ ] = 1.5 E-8 mol/L

Explanation:

  • CaF2 ↔ Ca2+  +  2F-

          S             S          2S......in the equilibrium

⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S )² = 4S³

⇒ S = ∛ ( 1.5 E-10 / 4 )

⇒ S = ∛ 3.75 E-11

⇒ S = 3.347 E-4 mol/L

⇒ [ Ca2+ ] = S = 3.347 E-4 mol/L

b) NaF ↔ Na+  +  F-

  0.10 M    0.10     0.10

  • CaF2 ↔,Ca2+  +    2F-

         S             S         2S + 0.10

⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S + 0.10 )²

∴∴ the Concentration: 0.10 M >>>> Ksp ( 1.5 E-10 ), son we can despise S as adding.

⇒ 1.5 E-10 = S * ( 0.10 )² = 0.01 S

⇒ S = 1.5 E-10 / 0-01

⇒ S = 1.5 E-8 mol/L

⇒ [ Ca2+ ] = S = 1.5 E-8 mol/L

5 0
3 years ago
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