Answer:
<em>The empirical formula is Ag2O.</em>
<em>The empirical formula is Ag2O.Explanation:</em>
<em>The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.</em>
<em>The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to </em><em>2</em><em>O.</em>
<em>do</em><em> </em><em>the</em><em> </em><em>steps</em><em> </em><em>.</em><em>.</em><em>.</em>
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
ElementAgMass/gXMolesXllRatiomllIntegers
mAgXXXm7.96Xm0.07377Xll2.00mmm2
mlOXXXXl0.59mm0.0369Xml1mmmml1
There are 2 mol of Ag for 1 mol of O.
Answer:
Yes, Temperature have an effect on capillary action.
Explanation:
increase in the temperature of liquid results to decrease in density. Hence, this results in the decrease in capillary rise. Also,increase in temperature leads to expansion of the material of capillary, this also decrease capillary rise.
Answer:
B = b -a/RT
C = b^2
a = 1.263 atm*L^2/mol^2
b = 0.03464 L/mol
Explanation:
In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:
Using the van deer Waals equation of state:
With further simplification, we have:
Then, we have:
Therefore,
Using the expansion:
Therefore,
Thus:
equation (1)
Using the virial equation of state:
Thus:
equation (2)
Comparing equations (1) and (2), we have:
B = b -a/RT
C = b^2
Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.
[/tex] = 0.03464 L/mol
a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2