When a female animal and a male see each other the male would show off and they woulf matr and make a baby then the babys will grow up and make more
Calculate the ratio by using Henderson-Hasselbalch equation:
pH = pKa + log [neutral form] / Protonated form
3.05 = 2.21 + log [neutral form] / [Protonated form]
3.05 - 2.21 = log [neutral form] / [Protonated form]
0.84 = log [neutral form] / [Protonated form]
[neutral form] / [protonated form] = anti log 0.84 = 6.91
The answer is <span>The components of a homogeneous mixture are evenly distributed.
In a homogeneous mixture, all components are evenly distributed. They are known as solutions. In a heterogeneous mixture, components are not evenly distributed. It consists of visibly different components. For example, milk is the homogeneous mixture, you cannot see its particles. But milk and cereals are the heterogeneous mixtures.</span>
<h2>Answer:</h2><h3>Part 1:</h3>
Location the element zinc (Zn) on the periodic table:
- Group number : 12
- Period number : 4
- Block : d block
- Element : Transition elements.
<h3>Part 2:</h3>
Protons in an atom of Zn: 30
<h3>Part 3:</h3>
Electrons in a Zn atom: 30
<h3>Part 4 :</h3>
Neutron in an atom of Zn: 35
<h3 />
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.
