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3 years ago

A compound is found to contain 73.23% xenon name 26.77% oxygen by mass. What is the empirical formula for this compound ?

Chemistry

1 answer:

Luba_88 [7]3 years ago

6 0

The empirical formula is XeO₃.

Explanation:

Assume 100 g of the compound is present. This changes the percents to grams:

Given mass in g:

Xenon = 73.23 g

Oxygen = 26.77 g

We have to convert it to moles.

Xe = 73.23/ 131.293 = 0.56 moles

O = 26.77/ 16 = 1.67 moles

Divide by the lowest value, seeking the smallest whole-number ratio:

Xe = 0.56/ 0.56 = 1

O = 1.67/ 0.56 = 2.9 ≈3

So the empirical formula is XeO₃.

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3 years ago

Read 2 more answers

Calculate the equilibrium constant K for the following reaction: H2(g) +

SIZIF [17.4K]

Answer:

192.9

Explanation:

From the question,

Ke = [HCL]²/[H₂][CL₂].......................... Equation 1

Where Ke = Equilibrium constant.

Given: [HCL] = 0.0625 M, [H₂] = 0.0045 M, [CL₂] = 0.0045 M

Substitute these values into equation 1

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3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

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Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.