The mass percent of oxygen is 32.82%.
<h3>What is a noble gas?</h3>
A noble gas is member of group 18 in the periodic table. The members of this group are known not to be reactive and they do not easily form compounds. However, there have been few compounds of the members of group 18 that have been reported.
Let us now calculate the relative molecular mass of xenon tetraoxide.
Xe + 4 (O)
131 + 4(16) = 195
Given that the mass of oxygen in the compound is 64, the mass percent of oxygen in xenon tetroxide is obtained from;
64/195 * 100/1
= 32.82%
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The primary colors are blue, red, and yellow.
Answer is: concentration ammonia is higher than concentration of ammonium ion.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,8 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,8 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² / 0,8 mol/L - x.
solve quadratic equation: x = c(NH₄⁺) = 3,79·10⁻³ mol/L.
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.