Explanation:
(a) The given data is as follows.
Load applied (P) = 1000 kg
Indentation produced (d) = 2.50 mm
BHI diameter (D) = 10 mm
Expression for Brinell Hardness is as follows.
HB = ![\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}](https://tex.z-dn.net/?f=%5Cfrac%7B2P%7D%7B%5Cpi%20D%20%5BD%20-%20%5Csqrt%7B%28D%5E%7B2%7D%20-%20d%5E%7B2%7D%29%7D%5D%7D)
Now, putting the given values into the above formula as follows.
HB = = ![\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%5Ctimes%201000%20kg%7D%7B3.14%20%5Ctimes%2010%20mm%20%5BD%20-%20%5Csqrt%7B%28%2810%20mm%29%5E%7B2%7D%20-%20%282.50%29%5E%7B2%7D%29%7D%5D%7D)
= 
= 200
Therefore, the Brinell HArdness is 200.
(b) The given data is as follows.
Brinell Hardness = 300
Load (P) = 500 kg
BHI diameter (D) = 10 mm
Indentation produced (d) = ?
d = ![\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}](https://tex.z-dn.net/?f=%5Csqrt%7B%28D%5E%7B2%7D%20-%20%5BD%20-%20%5Cfrac%7B2P%7D%7BHB%7D%20%5Cpi%20D%5D%5E%7B2%7D%29%7D)
= ![\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%2810%20mm%29%5E%7B2%7D%20-%20%5B10%20mm%20-%20%5Cfrac%7B2%20%5Ctimes%20500%20kg%7D%7B300%20%5Ctimes%203.14%20%5Ctimes%2010%20mm%7D%5D%5E%7B2%7D%7D)
= 4.46 mm
Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.
Larger molecules experience larger dispersion forces due to more distance of valance of electrons from the nucleus.
<h2>Cause of stronger dispersion force</h2>
Larger and heavier atoms and molecules have stronger dispersion forces than smaller and lighter ones because in a larger atom or molecule, the valence electrons are farther from the nuclei than in a smaller atom or molecule.
They are less tightly held to the nuclear charge present in the nucleus and can easily form temporary dipoles so we can conclude that larger molecules experience larger dispersion forces due to more distance of valance of electrons from the nucleus.
Learn more about London dispersion force here: brainly.com/question/1454795
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A gauge records the pressure over atmospheric pressure (0kpa on the gauge is actually the atmospheric pressure and a reading of 276kpa is 276kpa over atmospheric pressure). That means that means that to find absolute pressure you just add atmospheric pressure (around 1atm (101kpa)) to 286kpa to get 387kpa. I hope this helps.
Weak base: [OH⁻] = √Kb.C
pKb = 4.2
c = concentration
MM Amphetamine (C9H13N) = 135.21 g/mol
c = 215 mg/L = (0.215 g : 135,21 g/mol) / L = 0.00159 mol/L = 1.59 x 10⁻³ mol/L
![\tt [OH^-]=\sqrt{10^{-4.2}\times 1.59\times 10^{-3}}=3.17\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7B10%5E%7B-4.2%7D%5Ctimes%201.59%5Ctimes%2010%5E%7B-3%7D%7D%3D3.17%5Ctimes%2010%5E%7B-4%7D)
pOH = 4 - log 3.17
pH = 14 - (4 - log 3.17)
pH = 10 + log 3.17 = 10.50
