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Lelechka [254]
4 years ago
9

We learned about three career opportunities in the field of pulp and paper in class. Pick one career and find out the education,

training, and experience required for this profession. Write about what you found and explain why this profession might interest you
Chemistry
1 answer:
nekit [7.7K]4 years ago
8 0

Answer:

Pulp and Paper technology is a specialist field of chemical engineering which involves the study of the processes required for the conversion of raw materials such as wood, into pulp and paper products.

Paper and pulp manufacturing is a growing and dynamic industry so it requires a wide variety of personnel - from unskilled and semi-skilled workers to tradesmen, technicians and engineers - to control all the processes and develop new ones.

I choose to be a chemist in the field of paper and pulp. Requirements and compulsory Subjects: Physical Sciences, Mathematics, Engiineering and Technology.

Chemists are responsible for quality control and they study the influence of various chemicals on pulp and paper. Chemical and mechanical engineers develop new production methods.

I'll like to be a chemist in the field of paper and pulp because they perform the major role in the chemical properties of the paper and pulp.

Explanation:

Hope it helps.

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The relative humidity of the air is the ratio of the actual amount of moisture in the air to the fully saturated amount. There for the answer is 33.1% or 3.3g/kg
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Write 0.00103 in scientific notation
Andrei [34K]

Answer:

1.03

Explanation:

You would take 0.00103 and move the decimal like this; 0001.03, we wouldn't have the zeroes in front of the one, as it can throw us off. Therefore, your answer would be 1.03.

Hope I helped! Don't hesitate to let me know if I made a mistake.

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LolnnrjejahaushanakKababab
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Answer:

LolnnrjejahaushanakKababab

Explanation:

Source: Trust me bro

4 0
3 years ago
2.50 L of a gas at standard temperature and pressure is compressed to 575 mL. What is the new pressure of the gas
BartSMP [9]

Answer:

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Explanation:

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5 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
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