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Sveta_85 [38]
3 years ago
10

T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening,

the dispersed phase should be needle-like, as opposed to round___ Intermetallic compounds are usually hard and brittle.___ For effective dispersion strengthening, the dispersed phase should be continuous.___ Stoichiometric intermetallic compounds exist overa range of compositions.___ Faster solidification results in smaller interlamellar spacing
Chemistry
1 answer:
azamat3 years ago
4 0

Answer:

  • TRUE
  • FALSE
  • TRUE
  • FALSE
  • FALSE
  • TRUE

Explanation:

  • At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium
  • For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack.
  • Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hard
  • For the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous
  • Stoichiometric intermetallic compounds exist over a range of compositions : FALSE
  • Faster solidification results in smaller interlamellar spacing : TRUE
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So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

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Molarity of sulfuric acid solution = 0.5000 M

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Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

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  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

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So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

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Mass of metal = 1.00 g

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Putting values in above equation, we get:

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