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4 years ago

6

What are the three most common metals used in die casting?

1 answer:

Anit [1.1K]4 years ago

3 0

Answer:

Aluminium,Copper,Magnesium

Explanation:

The three most common metal of die casting are as follows

1.Aluminium

2.Copper

3.Magnesium

Die casting is the process in which metal is forced in the die to produces the desired casting product.Generally two type of machines are used like cold chamber and hot chamber machining,it depends on the metals.Die casting produces simple shape of casting ,it can not use for complex casting.

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Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel

guajiro [1.7K]

Answer:

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Explanation:

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2 years ago

What is the maximum thermal efficiency possible for a power cycle operating between 600P'c and 110°C? a). 47% b). 56% c). 63% d)

Anastasy [175]

Answer:

(b) 56%

Explanation:

the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature

here we have given T₁ (Higher temperature)= 600+273=873

lower temperature T₂=110+273=383

Efficiency of power cycle is given by =1- $\frac{T2}{T1}$

=1- $\frac{383}{873}$

=1-0.43871

=.56

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3 years ago

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Alex_Xolod [135]

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3 years ago

Convert A'B'C'D' + A'B'C'D + A'B'CD' + A'BC'D + AB'C'D' + AB'C'D+ AB'CD' to SOP form

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3 years ago

Read 2 more answers

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

4 0

3 years ago