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zavuch27 [327]
3 years ago
13

F has ___ valence electrons.

Chemistry
2 answers:
katrin [286]3 years ago
8 0
Fluorine is located 7 columns to the right on the periodic table. This means it has 7 valence electrons, and wants to gain 1 more in order to have the full valence shell of 8. 
Degger [83]3 years ago
8 0
There are 7 valence electrons in the element: Fluorine.
The electron configuration for Fluorine is:
1s2 ; 2s2 ; 2p7
There is a 7 at the end of 2p, indicating there are 7 valence electrons.
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Describe how a chemical change follows the law of conservation of matter.<br> give examples pls
Paha777 [63]

Explanation:

In any chemical change, one or more initial substances change into a different substance or substances. ... According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and kind of atoms after the chemical change as were present before the chemical change

Example:

The carbon atom in coal becomes carbon dioxide when it is burned. The carbon atom changes from a solid structure to a gas but its mass does not change.

5 0
3 years ago
What is true of all mass and volume of all floating objects
GarryVolchara [31]
If the density is higher than water than the object will sink
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It takes 330 joules of energy to raise the temperature of 24.6 gbenzene from 21 degrees Celsius to 28.7 degrees Celsius at const
lilavasa [31]

Given :

Energy , E = 330 J .

Initial temperature , T_i=21^oC .

Final temperature , T_f=24.6^oC .

Mass of benzene , m = 24.6 g .

To Find :

The molar hear capacity of benzene at constant pressure .

Solution :

Molecular mass of benzene , M = 78 g/mol .

Number of moles of benzene :

n=\dfrac{24.6}{78} \ mol\\\\n=0.32 \ mol

Energy required is given by :

q=nC_p\Delta T\\\\330=0.32\times C_p\times (28.7-21)\\\\C_p=\dfrac{330}{0.32\times 7.7}\ J\ mol^{-1}^oC^{-1} \\\\C_p=133.9\ J\ mol^{-1}^oC^{-1}

Hence , this is the required solution .

8 0
3 years ago
Consider a certain type of nucleus that has a half-life of 32 min. calculate the percent of original sample of nuclides remainin
Irina18 [472]
t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half life of the element and λ is decay constant.

32 = 0.693 / λ 
λ   = 0.693 / 32          (1) 

Nt = Nο eΛ(-λt)          (2)

Where Nt is atoms at t time, λ is decay constant and t is the time taken.
t = 1.9 hours = 1.9 x 60 min

From (1) and (2),


Nt = Nο e⁻Λ(0.693/32)*1.9*60
Nt =  0.085Nο 

Percentage = (Nt/Nο) x 100%
                   = (0.085Nο/Nο) x 100%
                   = 8.5%

Hence, Percentage of remaining atoms with the original sample is 8.5%

8 0
4 years ago
1. Which of the following equations is balanced?
pantera1 [17]
The answer would be A
6 0
3 years ago
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