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Shtirlitz [24]
2 years ago
11

Using data in Appendix A ,calculate the number of atoms in 1 tonne of iron

Engineering
1 answer:
maksim [4K]2 years ago
4 0
<h2>Answer:</h2>

1.0783*10^{28}(atoms).

<h2>Explanation:</h2>

<em>Since I don't have access to "Appendix A", I'll solve the problem using data from the periodic table.</em>

<em />

<h3>1. Determine the molar mass of iron.</h3>

<em>According to the periodic table, the molar mass of iron is:</em>

55.845g/mole.

<h3>2. Convert 1 tonne to grams.</h3>

1(tonne)*1000=1000kg\\1000kg*1000=1000000g=10^6g

<h3>3. Apply rule of 3.</h3>

55.845g ----------- 1 mole

10^6g ----------- x

x=\frac{10^6*1}{55.845}=17906.7061(moles)

<h3>4. Determine the amount of atoms.</h3>

<em>Considering that there are, approximately, </em>6.022*10^{23} atoms in a mole of any element, apply another rule of 3.

1 mole --------------------- 6.022*10^{23}(atoms)

17906.7061(moles) --------------------- x

x=\frac{17906.7061*6.022*10^{23}}{1}=1.0783*10^{28}.

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SB-4 Which one of the following is true about red buoys under the U.S. Aids to Navigation System?
stich3 [128]

Question:

1. Some are known as "nun" buoys

2. They are labeled with odd numbers

3. If it is lighted, the light color is green

4. Some are known as "can" buoys

Answer:

The correct option is;

1. Some are known as "nun" buoys

Explanation:

Based on the lateral system, on the starboard side, one can find the red even numbered marks while the odd-numbered, green, marks are located on the port side of a channel such that the buoy numbers increase as a vessel travels upstream.

The red buoys are cones shaped in appearance and have triangular reflective sign markings embossed and they are of different types included in the order of lower water depth

1. NUN buoy

2. Lighted buoy

3. Light

4. Day beacon

Therefore, the correct option is 1. Some are known as "NUN" buoys.

8 0
4 years ago
How might constraints and trade-offs affect the final design of a bridge?
Rainbow [258]

Answer:

It will limit the materials for your bridge which could cause damage to your bridge

Explanation:

4 0
3 years ago
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5 0
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If p = .8 and n = 50, then we can conclude that the sampling distribution of pˆ p ^ is approximately a normal distribution.
defon

Answer:

It is true that the sampling distribution of the proportions is approximately a normal distribution.

Explanation:

Solution

Given that:

N = 50

P= 0.8

Thus

p ^ =√p(1-p)/n

p ^=√0.8 (1-0.8)/50

p ^= 0.0566

Therefore,we conclude that the sampling distribution of the proportions is approximately a normal distribution.

4 0
3 years ago
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
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