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Nataly_w [17]
1 year ago
12

An element forms an oxide, E₂O₃, and a fluoride, EF₃.(b) How does the group to which E belongs affect the properties of the oxid

e and the fluoride?
Chemistry
1 answer:
xz_007 [3.2K]1 year ago
6 0

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.

To know more about Fluoride please click here : brainly.com/question/10929330

#SPJ4

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anyanavicka [17]

Answer:  The percent ionization of CH_3NH_2 in a 0.050 M CH_3NH_2(aq) solution is 8.9 %

Explanation:

CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+

 cM                            0             0

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So dissociation constant will be:

K_b=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= concentration = 0.050 M and \alpha = degree of ionisation = ?

K_b=4.4\times 10^{-4}

Putting in the values we get:

4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}

(\alpha)=0.089

percent ionisation =0.089\times 100=8.9\%

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