Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
The major visible difference between<span> the two are crystal size, </span>intrusive rocks<span> have a larger crystal/grain texture due to the slow cooling of magma below the earth surface which encourages the growth of larger crystals, while </span>extrusive rocks<span>, because of the rapid cooling at/above the earth's surface does the opposite. Hope I helped</span>
Answer:
Explanation:
Given that
d= 35 μm ,yield strength = 163 MPa
d= 17 μm ,yield strength = 192 MPa
As we know that relationship between diameter and yield strength


d = diameter
K =Constant

So now by putting the values
d= 35 μm ,yield strength = 163 MPa
------------1
d= 17 μm ,yield strength = 192 MPa
------------2
From equation 1 and 2

K=394.53
By putting the values of K in equation 1


Now when d= 12 μm

