Students are completing an investigation on types of heat transfer. For one part of the investigation, they place their hand on
1 answer:
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Explanation:
The given precipitation reaction will be as follows.
Here, AgCl is the precipitate which is formed.
It is known that molarity is the number of moles present in a liter of solution.
Mathematically, Molarity =
It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.
Molarity =
0.269 M =
no. of moles = 0.306 mol
As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.
No. of moles =
0.307 mol =
mass = 43.99 g
Thus, we can conclude that mass of precipitate produced is 43.99 g.
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 =
Now, putting the given values into the above formula as follows.
HB = =
=
= 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 =
=
= 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.