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3 years ago

Write down one metal or alloy that is best suited for each of the following applications:

Engineering

1 answer:

777dan777 [17]3 years ago

6 0

Answer:

Explanation:

a. Cast iron or Aluminium alloy are typically used. Aluminium is much lighter in weight and it can transfer heat better to the coolant. While Cast Iron is typically stronger and is thus still used by the manufacturers.

b. Copper can be used as a condensing heat exchanger for hot steam due to its optimal thermal properties and its ability to resist corrosion.

c. high-speed steel are perfect for producing drill bits because of its hardness and resistance to heat to an extent. Drill bits tend to produce heat as a result of the friction between it and the material to be drilled.

d. lead can be used as a container for strong acids because of its anti-corrosive properties

e.zinc and copper can be used as fuel in pyrotechnics mainly due to the fact that burn with refreshing colours. Aluminium can also be used.

f. Platinum is the metal that best suits this purpose because of its high melting point and resistivity to oxidation.

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2 years ago

A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m

natima [27]

Answer:

a. 0.4544 N

b. $5.112 \times 10^{-5 M}$

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

$H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O$

$NaOH\ Mass = Normality \times equivalent\ weight \times\ volume$

$= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL$

= 0.27264 g

$NaOH\ mass = \frac{mass}{molecular\ weight}$

$= \frac{0.27264\ g}{40g/mol}$

= 0.006816 mol

Now

Moles of $H_2SO_4$ needed is

$= \frac{0.006816}{2}$

= 0.003408 mol

$Mass\ of\ H_2SO_4 = moles \times molecular\ weight$

$= 0.003408\ mol \times 98g/mol$

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

$= \frac{acid\ mass}{equivalent\ weight \times volume}$

$= \frac{0.333984 g}{49 \times 0.015}$

= 0.4544 N

b. And, the acid solution molarity is

$= \frac{moles}{Volume}$

$= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}$

= 0.00005112

= $5.112 \times 10^{-5 M}$

We simply applied the above formulas

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3 years ago

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hoa [83]

Answer:

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