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Arlecino [84]
2 years ago
6

A cylindrical specimen of some metal alloy having an elastic modulus of 126 GPa and an original cross-sectional diameter of 3.4

mm will experience only elastic deformation when a tensile load of 2130 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Chemistry
1 answer:
Delvig [45]2 years ago
5 0

Answer:

The maximum length of the specimen before deformation is 240.64 mm

Explanation:

Strain = stress ÷ elastic modulus

stress = load ÷ area

load = 2130 N

diameter = 3.4 mm = 3.4×10^-3 m

area = πd^2/4 = 3.142 × (3.4×10^-3)^2/4 = 9.08038×10^-6 m^2

stress = 2130 N ÷ 9.08038×10^-6 m^2 = 2.35×10^8 N/m^2

elastic modulus = 126 GPa = 126×10^9 Pa

Strain = 2.35×10^8 ÷ 126×10^9 = 0.00187

Length = extension ÷ strain = 0.45 mm ÷ 0.00187 = 240.64 mm

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Answer:

The answer to your question is 88.7 ml

Explanation:

Data

Volume = ?

Concentration of NaOH = 0.142 M

Volume of H₂C₄H₄O₆ = 21.4 ml

Concentration of H₂C₄H₄O₆ = 0.294 M

Balanced chemical reaction

               2 NaOH + H₂C₄H₄O₆  ⇒  Na₂C₄H₄O₆  +  2H₂O

1.- Calculate the moles of H₂C₄H₄O₆

Molarity = moles/volume

Solve for moles

moles = Molarity x volume

Substitution

moles = 0.294 x 21.4/1000

Result

moles = 0.0063

2.- Use proportions to calculate the moles of NaOH

              2 moles of NaOH ------------------ 1 moles of H₂C₄H₄O₆

               x                           ------------------ 0.0063 moles

               x = (0.0063 x 2) / 1

               x = 0.0126 moles of NaOH

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Molarity = moles / volume

Solve for volume

Volume = moles/Molarity

Substitution

Volume = 0.0126/0.142

Result

Volume = 0.088 L or 88.7 ml

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elena-14-01-66 [18.8K]
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