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mixas84 [53]
3 years ago
14

Give the electron configuration for F.

Chemistry
1 answer:
postnew [5]3 years ago
4 0

Answer:

1s² 2s² 2p⁵

Explanation:

The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements can not be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements. Their boiling points also increases down the group which changes their physical states. i.e fluorine is gas while iodine is solid.

Properties of fluorine:

1. it is yellow in color.

2. it is flammable gas.

3. it is highly corrosive.

4. fluorine has pungent smell.

5. its reactions with all other elements are very vigorous except neon, oxygen, krypton and helium.  

6. Its atomic number of 9.

7. its atomic mass is 18.99 amu.

Electronic configuration:

₉F = 1s² 2s² 2p⁵

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Answer: True

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3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
Of the five salts listed below, which has the highest concentration of its cation in water? assume that all salt solutions are s
Zigmanuir [339]
First of all, I need to know what these five salts are. Luckily, I found a similar problem from another website which is shown in the attached picture. The Ksp is the solubility product constant. It follows the formula:

Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively. 

For the solutions ahead, let x be the concentration of the cation.

A.  The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M

B. The formula is Co(OH)₂. 
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Solving for x, x = 1.09×10⁻⁵ M

C. The formula is CoS. 
 5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M

D. The formula is Cr(OH)₃. 
 1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M

E. The formula is Ag₂S. 
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<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>

5 0
3 years ago
If this is a p1000 micropipette, then this is set to dispense [ Select]ul. If this is a p10
mariarad [96]

Answer:

1000 µL; 10 µL  

Explanation:

A p1000 micropipet is set to dispense 1000 µL.

A p10 micropipet set to dispense 10 µL.

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How is liming a lake similar to a doctor prescribing medicine for a patient?
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