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Sonja [21]
3 years ago
7

How many milliliters of 0.183 m hcl would be required to titrate 5.93 g koh?

Chemistry
1 answer:
jek_recluse [69]3 years ago
4 0
  Molar mass:
KOH = 56.0 g/mol

Number of moles of KOH :

5.93  / 36.5 => 0.1624 moles

<span>KOH + HCl = KCl + H₂O 
</span>
1 mole KOH --------------> 1 mole  HCl
0.1624 moles KOH ----> ?

moles HCl = 0.1624 x 1 / 1

= 0.1624 moles of HCl

V ( HCl ) = moles / molarity

V(HCl) = 0.1624 / 0.183

V (HCl) = 0.887 L x 1000 = 887 mL

hope this helps!



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                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

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               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

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                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 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.

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