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4 years ago

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Gold forms a substitutional solid solution with silver. Calculate the number of gold atoms per cubic centimeter (in atoms/cm3) f

or a silver-gold alloy that contains 17 wt% Au and 83 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3, respectively, and their respective atomic weights are 196.97 and 107.87 g/mol. atoms/cm3

1 answer:

bija089 [108]4 years ago

8 0

Answer:

Na = 5.911 × $10^{21}$ atoms/cm³

Explanation:

given data

silver-gold alloy C Au = 17 wt%

densities ρ Au = 19.32 g/cm³

densities ρ Ag = 10.49 g/cm³

atomic weights A Au = 196.97 g/mol

atomic weights A Ag = 107.87 g/mol

solution

we will apply here formula for number of gold atoms that is

Na = $\frac{NA * C Au}{\frac{C Au * A Au}{\rho Au} + \frac{A Au * (100-C Au)}{\rho Ag}}$ ............................1

here NA is Avogadro's number and ρ is density of two element and A is atomic weight

put here value

Na = $\frac{6.022*10^{23} * 0.17}{\frac{0.17 * 196.97}{19.32} + \frac{196.97 * (100-0.17)}{10.49}}$

Na = 5.911 × $10^{21}$ atoms/cm³

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Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manome

oksano4ka [1.4K]

Answer:

a) Δh = 2 cm, b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1

y₁ = y₂

subtitute

P₁ = P₂ + ½ ρ v₂²

P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

ΔP =ρ_{Hg} g h - ρ g h

ΔP = (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

ΔP = P₁ - P₂

(ρ_{Hg} - ρ) g h = ½ ρ v²

v = $\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}$

in this expression the densities are constant

v = A √h

A = $\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }$

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

A = $\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }$

A = 454.55

we substitute

v = 454.55 √h

to calculate the uncertainty or error of the velocity

h = $\frac{1}{454.55^2} \ v^2$

Δh = $\frac{dh}{dv}$ Δv

$\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}$

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

$\frac{\delta v}{v} = 0.05$

$\frac{\Delta h}{h}$ = 2 0.05

Δh = 0.1 h

Δh = 0.1 20 cm

Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

$\frac{\Delta h}{h}$ = 2 0.01

Δh = 0.02 h

Δh = 0.02 20

Δh = 0.1 20 cm

Δh = 0.4 cm = 4 mm

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3 years ago

Your department's laser printer recently began printing a vertical black line near the edge of every printed page. What should y

WINSTONCH [101]

Answer:

Replace the toner cartridge

Explanation:

solution

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but first we clean the corona wire for that color line

and Reinstall the toner cartridges after Shake the toner cartridge side to side

if problem not solve than Clean the drum unit

and finally if not solve change the toner cartridge

so as that our problem will be resolve

5 0

3 years ago

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500

lukranit [14]

Answer:

The resulting strain is $1.39\times 10^{-3}$ .

Explanation:

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm

Force, F = 44,500 N

Th elastic modulus of Cu to be 110 GPa

The resulting strain is given by the formula as follows :

$\epsilon=\dfrac{F}{AE}$

E is elastic modulus of Cu is are of cross section

$\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}$

So, the resulting strain is $1.39\times 10^{-3}$ .

6 0

4 years ago

In a balanced Y-Y three-phase circuit 120 10 Van    V rms. The load impedance per phase is 20 15 ZZ j L Y    . Determine

ANEK [815]

Given Information:

Balanced Y-Y three phase circuit

Phase Voltage = Van = 120 < 10° V

Load Impedance = Zy = 20 +j15 Ω = 25 < 36.86° Ω

Required Information:

Load Voltages = Vab, Vbc, Vca = ?

Line currents = Ia, Ib, Ic = ?

Answer:

Vab = 208 < 40°

Vbc = 208 < -80°

Vca = 208 < 160°

Ia = 4.8 < -26.86°

Ib = 4.8 < -146.86°

Ic = 4.8 < 93.14°

Explanation:

Since it is balanced 3 phase system, all phases have equal magnitude and 120° phase shift

Van = 120 < 10° V

Vbn = 120 < -110° V

Vcn = 120 < 130° V

In a Y connected system, the phase voltage and line voltage are related as

Vab = $\sqrt{3}$ (Van) <+30°

Vab = $\sqrt{3}$ *120<10°+30°

Vab = 208 < 40°

Vbc = 208 < -80°

Vca = 208 < 160°

In a Y connected circuit, Iphase = Iline

Ia = Van/Zy

Ia = 120 < 10°/25 < 36.86°

Ia = 4.8 < -26.86°

Ib = 4.8 < -146.86°

Ic = 4.8 < 93.14°

7 0

4 years ago

Consider a thin-walled cylindrical tube having a radius of 65 mm that is to be used to transport pressurized gas. (a) (10 points

AlladinOne [14]

Answer:

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

Explanation:

Minimum required thickness is the thickness of a material without corrosion allowance for each component based on the appropriate design that consider pressure, mechanical and structural loading.

Given that:

radius (r) = 65 mm = 65 × 10⁻³ m

Factor of safety (N) = 3.5

Inside presssure ( $P_{in}$ ) = 11 MPa

Outside pressure ( $P_{out}$ ) = 1 MPa

Yield strength ( $\sigma_y$ ) = 1000 MPa

Therefore:

$\sigma_y =\frac{\sigma_y}{N}$ , Substituting values,

$\sigma_y =\frac{\sigma_y}{N}=\frac{1000}{3.5}=285.714 MPa$

The minimum required thickness for a steel pressure vessel (t) is given by the equation:

$t=\frac{r.(P_{in}-P_{out})}{\sigma_y}$ . Substituting values

$t=\frac{r.(P_{in}-P_{out})}{\sigma_y}=\frac{65*10^{-3} *(11-1)10^{6} }{285.714*10^{6} } =2.275*10^{-3} =2.275 mm$

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

7 0

3 years ago