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cluponka [151]
3 years ago
8

A compound die will be used to blank and punch a large washer out of aluminum alloy sheet stock 3.2 mm thick. the outside diamet

er of the washer = 65 mm and the inside diameter = 30 mm. determine: (a) the punch and die sizes for the blanking operation, and (b) the punch and die sizes for the punching operation.
Chemistry
1 answer:
rodikova [14]3 years ago
7 0

a) The sizes of punch and blanking are equivalent to its diameters. The blanking die diameter is Db = 65 mm

Diameter of the blanking punch is,

dpunch = Db - 2c

Here, c = Ac (thickness)

For aluminium, Ac = 0.06

c = (0.06) (3.2 mm) = 0.192 mm

dpunch = (65 mm) - 2 (0.192 mm)

= 64.616 mm

b) The diameter of the punching punch is equivalent to the interior diameter of the washer.

Thus, diameter of the punching die is, Dh = 30 mm

Diameter of the punching die is,

ddie = Dh + 2c

= (30 mm) + 2 (0.192)

= 30.384 mm

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Answer: Option (E) is the correct answer.

Explanation:

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For example, ice melting at 25^{o}C is spontaneous primarily due to the increase in molecular disorder (dispersal of matter). Also, melting of ice is taking place on its own without any external force.

It is not necessary that all exothermic reactions will be exothermic in nature.

Thus, we can conclude that the statement all exothermic reactions are spontaneous, is false.

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Which of the following would increase the amount to dissolve a solid solute?
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By stirring and increasing temperature, there is an increase in dissolving capacity of the solid solute.

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If a solute is added to the solution, it doesn't get dissolve easily then we have to increase the temperature, which in turn increases the movement of the solvent (may be water) and the solute particles, thus increases the dissolving power of the solid solute. One more way is by constant stirring, that is by making more contact among the solvent as well as the solute particles there by increasing the solubility of solid solute.

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A car is moving with the velocity of 20m/s. After 5 seconds it's velocity becomes 50m/s. Find its acceleration.​
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Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n
mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide (CO_2), nitrogen dioxide (NO_2), and no other substances. A small sample gives 2.389 g CaO, 1.876 g CO_2, and 3.921 g NO_2 Determine the empirical formula of the compound.

<u>Answer:</u> The empirical formula for the given compound is CaCN_2

<u>Explanation:</u>

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of CO_2=1.876g

Mass of NO_2=3.921g

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, \frac{12}{44}\times 1.876=0.5116g of carbon will be contained.

<u>For calculating the mass of nitrogen:</u>

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, \frac{14}{46}\times 3.921=1.193g of nitrogen will be contained.

<u>For calculating the mass of calcium:</u>

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, \frac{40}{56}\times 2.389=1.706g of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Calcium =\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles

Moles of Nitrogen = \frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = \frac{0.0426}{0.0426}=1

For Carbon = \frac{0.0426}{0.0426}=1

For Nitrogen = \frac{0.0852}{0.0426}=2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is CaCN_2

3 0
3 years ago
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