Minerals are solid, naturally occurring, inorganic compounds that possess an orderly internal structure and a regular chemical composition. Minerals should occur naturally. Hope this answers the question. Have a nice day. Feel free to ask more questions.
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.
Addition of boiled, deionized water to the titrating flask to wash the wall of the erlenmeyer flask and the buret tip will have no effect on the Ksp value of ca(oh)2.
There will be no effect on the Ksp value as boiled deionised water is not able to alter the number of hydronium and hydroxide ions. As no change in the ions happen so there will be no change in Ksp value. The equilibrium constant for a solid material dissolving in an aqueous solution is the solubility product constant, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.
To know more about, solubility product constant, click here,
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Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) = ![\frac{ΔvapH}{R}([tex]\frac{1}{T1}](https://tex.z-dn.net/?f=%5Cfrac%7B%CE%94vapH%7D%7BR%7D%28%5Btex%5D%5Cfrac%7B1%7D%7BT1%7D)
-
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ
vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P
1
=
100 mmHg
; T
1
=
34.7 °C
=
307.07 K
P
2
=
760mmHg
T
2
=T⁻²=?
Δ
vap
H
=
38.6 kJ/mol
R
=
0.008314 kJ⋅K
-1
mol
-1
ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K
