Fluorine has C seven valence electrons.
Valence electrons are the electrons that are found on the outer shell of an atom that are capable of participating in chemical reactions. The easiest way to figure out how many valance electrons Fluorine has would be to look it up in a periodic table and notice that it a group 7 element and therefore has 7 valence electrons.
The other way to tell that Fluorine has 7 valence electrons is to notice that an element with an atomic number of 9 has 9 electrons. The electronic configuration of elements has the first 2 electrons go on the first shell or energy level 
, the leaving the next 7 to go to the second shell 
which can take up to 8 electrons.
This means that Fluorine has 7 seven valence electrons.
Answer: All of the statements (A-D) are false
Explanation: All of the given statements are wrong as Hydrogen bonds are weaker than the ionic and the covalent bonds.
Liquid water is more dense than the solid water.
When heat or temperature is provided to the system of the ice, not only covalent bonds are broken but also hydrogen bonds also get broke.
Energy must not be given off in order to break down the crytsal lattice of ice to a liquid. Infact in changing the phase of the water, energy is released.
Valence electrons hope this helps
Answer:
The answer to this is
The velocity of the 27.3Kg marble after collision is = 16.24 cm/s
Explanation:
To solve the question, let us list out the given variables and their values
Mass of first marble m1 = 27.3g
Velocity of the first marble v1 = 21.0 cm/s
Mass of second marble m2 = 11.7g
Velocity of the second marble v2 = 12.6 cm/s
After collision va1 = unknown and va2 = 23.7 cm/s
From Newton's second law of motion, force = rate of change of momentum produced
Hence m1v1 + m2v2 = m1va1 + m2va2 or
va1 = (m1v1 + m2v2 - m2va2)÷m2 or (720. 72-277.29)÷m1 → va1 = 16.24 cm/s
The velocity of the 27.3Kg marble after collision is = 16.24 cm/s
<span>Given:
acid-dissociation constants of sulfurous acid</span>:
Ka1 = 1.7 * 10^(-2)
Ka2 = 6.4 * 10^(-8) at
25.0 °C.
aqueous solution of
sulfurous acid = 0.163 M
x² / (0.163 - x) = 1.7 * 10^(-2)
You simplify it to:
<span>x² / (0.163) = 1.7 *10^(-2) </span>
x = 0.052640 M
pH = 1.28
<span>
So, the pH of a 0.163 M aqueous solution of sulfurous acid is 1.28.</span>
To add, aqueous solutions of sulfur dioxide purpose
are as disinfectants and reductant, as are solutions of sulfite<span> salts
and </span>bisulfite. By accepting another oxygen<span> atom, they
are </span>oxidised to sulfuric
acid or sulfate.
