Of all the elements, fluorine is the most electronegative and reactive. Fluorine is a diatomic, pale yellow, extremely corrosive, combustible gas with a strong smell. The lightest halogen is it. It produces oxygen and the incredibly corrosive hydrofluoric acid when it combines strongly with water.
<h3>The properties of the oxide and the fluoride?</h3>
- 1. A mixture of oxygen fluorides with an atomic ratio OF in the range of 1.1-2.04 is generated when fluorine and oxygen mixes are easily circulated through a silent electric discharge.
- Depending on where you reside in the UK, fluoride is a naturally occurring mineral that is present in water in variable concentrations. It is added to many types of toothpaste and, in some locations, the water supply through a procedure known as fluoridation because it can aid in the prevention of tooth decay.
- Fluoride stops tooth decay by strengthening the enamel's resistance to acid attack. They also quicken the process of good minerals accumulating in the enamel, further delaying the onset of deterioration. Studies also suggest that fluoride may occasionally be able to stop tooth decay that has already begun.
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Oxygen : 367*0.888=325.896
Hydrogen : 367 - 367*0.888 = 41.104g
Answer:
oxidation state of sulphur=x
Explanation:
Na2S4O6=2[+1]+4x+6[-2]=0
+2+4x-12=0
4x-10=0
4x=10
x=10/4=2.5
It is easier to determine this if we draw the structural formula of lactic acid as shown in the attached picture. There are three functional groups in lactic acid: carboxyl group, hydroxyl group, and the parent alkane chain. Any of the hydrogens in the alkane chain is the least acidic. Then, it is followed by the H in the hydroxyl group. The most acidic is the H in the carboxyl group.
Answer:
%Ionization = 1.63%
Explanation:
Hydrazine in aqueous media theoretically forms a difunctional hydroxyl system. However, for this problem assume only monofunctional ionization occurs. A second hydroxyl ionization would not likely occur as the formal cationic charge formed in the 1st ionization would inhibit a second ionization.
H₂NNH₂ + 2H₂O => HONHNHOH => HONHNH⁺ + OH⁻; Kb = 1.3 x 10⁻⁶
So, assuming all OH⁻ and HONHNH⁺ are delivered in the 1st ionization then a good estimate of the %ionization can be calculated.
HONHNHOH => HONHNH⁺ + OH⁻
C(i) => 0.490M 0M 0M
ΔC => -x +x +x
C(eq) => 0.490 - x x x
≅0.490M* => *x is dropped as Conc H₂NNH₂/Kb > 100
Kb = [HONHNH⁺][OH⁻]/[HONHNHOH]
1.3 x 10⁻⁶ = x²/0.490
=> x = [OH⁻] = [HONHNH⁺] = √[(1.3 x 10⁻⁶)(0.490)] = 8 x 10⁻⁴
=> %Ionization = (x/0.490)100% = (8 x 10⁻⁴/0.490)100% = 1.63%
