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.
Answer:
Compound B would have stronger bonds because the reaction needs and absorbs heat to reach the status of the product.
Explanation:
Answer:
Explanation:
The term used by particular kind of matter is called substance.
A substance is a particular kind of matter because it has physical properties.
Answer:
No. While gold would not react with a silver nitrate solution, nickel would.
Explanation:
Refer to the metal reactivity series.
Reactivity: 
.
Gold is positioned after silver in the reactivity series, meaning that gold is typically less reactive than silver. Thus, gold would not react with a solution of silver ions to produce silver metal.
However, since nickel is positioned before silver in the reactivity series, it is expected that nickel would react with silver ions in this solution to produce silver metal.
Thus, if the silver nitrate solution comes into contact with the two rings, the nickel ring would likely react with the solution, the gold ring would not.
The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.
If 1 mole of vinegar contains 6.02 x 10^23 particles
x moles of vinegar contains 9.02 x 10^24 particles
x = 1 mole x 9.02 x 10^24 /6.02 x 10^23
x = 15 moles of vinegar
The reaction is as follows;
2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2
Since 2 moles of vinegar reacts with 1 mole of carbonate
x moles of vinegar reacts with 16.5 moles of carbonate
x = 2 moles x 16.5 moles/ 1 mole
x = 33 moles of vinegar
We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.
Theoretical yield = 16.5 moles x 158 g/mol = 2607 g
Actual yield = 6.35 moles x 158 g/mol = 1066.8 g
Percent yield = 1066.8 g/2607 g × 100/1
= 41%
Learn more: brainly.com/question/13440572?
