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:
combustion is a chemical change
Given :
10 gram sample of propane( C₂H₈ ).
To Find :
The number of moles of propane in given sample.
Solution :
Molecular mass of propane, M = (2 × 12) + ( 1 × 8 ) gram/mol
M = 32 gram/mol
We know, number of moles is given by :
Number of moles, n = m/M
n = 10/32 mol
n = 0.3125 mol
Therefore, number of moles in given sample is 0.3125 mol.
Answer:
44. (C) The temperature must decrease because the reaction is endothermic
45. (D) 3Pi
46. (C) It must be positive since positive since ΔG° is positive and ΔH° is positive
47. (B) The sum of the bond enthalpies of the bonds of the reactant is less than the sum of the bond enthalpies of the bonds of the products
Explanation:
Here we have
CH₃OH(g) → CO(g) + 2H₂(g) ΔH° +91 kJ/mol
44. Since the reaction is endothermic, absorbs heat, temperature must decrease because the reaction is endothermic
45. Since the number of moles in the reactant is 1 and the number of moles in the product is 3, we have;
Pressure, P is directly proportional to the number of moles
Therefore, where the pressure in the reactant is Pi pressure in the products will be 3Pi
46 Since the reaction takes place spontaneously at 600 K, therefore ΔG is negative and ΔH is positive hence ΔS must be positive
47. Since the reaction is an endothermic reaction, the sum of the bond enthalpies of the bonds of the reactant is less than the sum of the bond enthalpies of the bonds of the products.
