The empirical formula is XeO₃.
<u>Explanation:</u>
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.
Let us assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2
heat absorbed by pork = 0.11 x 2217
= 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
= 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
= 44 x 20.913
= 920.172g of CO2 emitted
Answer:
The half-life of a radioisotope describes the amount of time it takes for said isotope to decay to one-half the original amount present in the sample.
Nitrogen-13, because it has a half-life of ten minutes, will experience two half-lives over the course of the twenty minute period. This means that 25% of the isotope will remain after this.
0.25 x 128mg = 32mg
32mg of Nitrogen-13 will remain after 20 minutes.
Answer:
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O
Explanation:
NO₃⁻ → NO₂
In left side, Nitrogen acts with +5 by oxidation number
In right side, the oxidation number is +4
This is a reduction reaction, because the oxidation number has decreased. So the N has gained electrons.
NO₃⁻ + 1e⁻ → NO₂
In acidic medium, we have to add water, where there are less oxygens to ballance the amount. We have 2 O in left side, and 3 O in right side, so we have to add 1 H₂O on left side.
NO₃⁻ + 1e⁻ → NO₂ + H₂O
Now that oxygens are ballanced, we have to ballance the hydrogens by adding protons in the opposite side
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
