<span>Van der waal or ideal eqn is given by PV = NRT; P = NRT/ V.
Where N = 1.335 is the number of moles. T = 272K is temperature. V = 4.920L is the volume. And R = 0.08205L. Substiting the values into the eqn; we have,
P = (1.331* 0.08205 * 272)/ 4.920 = 29.7047/ 4.920 = 6.03atm.</span>
Sorry I’m only answering so I could upload
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
Answer:
A) M = 100X
B) M = 36X
C) M = 178.88X
Explanation:
Given data:
ASTM grain size number 7
a) total grain per inch^2 - 64 grain/inch^2
we know that number of grain per square inch is given as
where M is magnification, n is grain size
therefore we have
solving for M we get
M = 100 X
B) total grain per inch^2 = 500 grain/inch^2
we know that number of grain per square inch is given as
where M is magnification, n is grain size
therefore we have
solving for M we get
M = 36 X
C) Total grain per inch^2 = 20 grain/inch^2
we know that number of grain per square inch is given as
where M is magnification, n is grain size
therefore we have
solving for M we get
M = 178.88 X
