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sweet [91]
3 years ago
15

What’s the number of gold atoms in a nanogram? a picogram?

Engineering
1 answer:
zvonat [6]3 years ago
8 0

Answer :

The number of gold atoms in nanogram is, 3.057\times 10^{12}

The number of gold atoms in picogram is, 3.057\times 10^{9}

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains 6.022\times 10^{23} number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-9} nanograms of gold contains \frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12} number of gold atoms

The number of gold atoms in nanogram is, 3.057\times 10^{12}

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-12} picograms of gold contains \frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9} number of gold atoms

The number of gold atoms in picogram is, 3.057\times 10^{9}

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Answer:

<h2>True </h2>

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8 0
3 years ago
Using the AASHTO 1993 Flexible Pavement Design Procedure, design a pavement cross section that will provide 10 years service. Th
natali 33 [55]

Answer:

Check the explanation

Explanation:

Single Unit Truck ESAL = 43.38 + 5.16 = 48.54

Semi Unit Truck ESAL = 43.38+ 6.00+7.4 = 56.78

So total ESAL's during design life = (400*48.54 + 350*56.78)*365*10/18000 = (19416+19873)*3650= 3939

Kindly check the attached image

Here

Reliability = 95% = 0.95, therefore ZR = -1.645, S0 = 0.4, MR = 18

Delta PSI = 4.2-2.5= 1.7

Resilient Modulus = 18000 psi, So MR = 18

Assume SN = 3.0 for flexible pavements

There W18 calculates to 0.26807

So

log10 (3939) = 9.36*log10(SN+1) -.2/(.4+1094/(SN+1)5.19)) -6.01

Structural Number SN = a1*d1 + a2*d2 *m2 +a2*d3 *m3

= a1*d1 + a2*d2 +a2*d3

5 0
3 years ago
The diffusion coefficients for iron in nickel are given at two temperatures:T (K)D (m2/s)12739.4 × 10–1614732.4 × 10–14(a) Deter
hram777 [196]

Answer:

The diffusion coefficients for iron in nickel are given at two temperatures:

T (K)        1273          1473

D (m^{2}/s) 9.4 × 10^{-16}    2.4 × 10^{-14}

(a) Determine the values of Do and the activation energy Qd.

(b) What is the magnitude of D at 1100°C (1373 K)?

<em>A </em>

<em>The pre-exponential factor Do = 2.1 x </em>10^{-16}<em></em>

<em>The activation energy Qd = 252,609 J/mol</em>

<em>B</em>

<em>The diffusion coefficient D= 5.14 x </em>10^{-15}<em></em>

Explanation:

The full explanation is contained in the attached images;

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4 years ago
An automobile tire with a volume of 0.8 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if t
saveliy_v [14]

Answer:

2.83 kg

Explanation:

Given:

Volume, V = 0.8 m³

gage pressure, P = 200 kPa

Absolute pressure = gage pressure + Atmospheric pressure

= 200 + 101 = 301 kPa = 301 × 10³ N/m²

Temperature, T = 23° C = 23 + 273 = 296 K

Now,

From the ideal gas equation

PV = mRT

Where,

m is the mass

R is the ideal gas constant = 287 J/Kg K. (for air)

thus,

301 × 10³ × 0.8 = m × 287 × 296

or

m = 2.83 kg

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