Can you help me with this task/homework.

You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19

Pepsi [2]

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

$V = A*t$

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

$V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}$

Now, using the density we can find the mass:

$m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g$

Finally, with the Avogadros number ( $N_{A}$ ) and with the atomic mass (A) we can find the number of atoms (N):

$N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms$

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0

3 years ago

99 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

qwelly [4]

Answer:

1. Can you tell me something about yourself?

2. What are you weaknesses?

3. If you would describe yourself in one word?

Explanation: Those questions above 1, 2, and 3 are not harmful to ask your client. Bit the last two 4 and 5 are very harmful, because you don't need to be all up in they business and you don't want to put a lot of pressure on your client.

Hope this helps☝️☝☝

7 0

3 years ago

Read 2 more answers

Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a

Arada [10]

Answer:

a) $h_c = 0.1599 W/m^2-K$

b) $H_{loss} = 5.02 W$

c) $T_s = 302 K$

d) $\dot{Q} = 25.125 W$

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, $T_a = 25^0 C = 298 K$

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

$\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600} = 1.25 kg/sec$

Rate of heat transfer,

$\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W$

a) To calculate the convection coefficient relationship for heat transfer by convection:

$\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K$

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, $k_c = 0.8$

$\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K$

b) Heat loss from the pipe to the environment:

$H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W$

d) The required fan control power is 25.125 W as calculated earlier above

5 0

3 years ago

A gas cylinder is connected to a manometer that contains water. The other end of the manometer is open to the atmosphere, which

horrorfan [7]

Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa

Explanation:

Given that;

patm = 79 kPa, h = 13 in of H₂O,

A sketch of the problem is uploaded along this answer.

Now

pA = patm + 13 in of H₂O ( h × density × g )

pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)

pA = 82.23596 kPa

the absolute static pressure in the gas cylinder is 82.23596 kPa

4 0

3 years ago

: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the

Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

$\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}$

$V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s$

$\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}$

taking f = 0.0185

at Z₁ = Z₂

$\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}$

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P= 1.13\ psi$

c) When flow is downhill z₂-z₁ = -2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P=-0.601\ psi$

7 0

3 years ago