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torisob [31]
3 years ago
9

What do you think causes the differences in the properties of oxygen gas (O2) and ozone (O3)? the space between the atoms the ra

tio of atoms in the molecules cannot tell the number of oxygen atoms
Engineering
1 answer:
Ivahew [28]3 years ago
4 0

Answer:

Oxygen is an important part of the atmosphere and is necessary to sustain terrestrial life. Because it comprises most of the mass in water, it also comprises most of the mass of living organisms. All major classes of structural molecules in living organisms, such as proteins, carbohydrates, and fats, contain oxygen, as do the major inorganic compounds that comprise animal shells, teeth, and bone. Elemental oxygen (O2) is produced by cyanobacteria, algae, and plants through the process of photosynthesis, and is used in cellular respiration by most living organisms on earth. Oxygen is toxic to obligate anaerobic organisms (organisms which need a lack of oxygen for survival), which were the dominant form of early life on Earth, until O2 began to accumulate in the atmosphere.

Explanation:

You might be interested in
g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70%
Bond [772]

Answer:

<em>Heat rejected to cold body = 3.81 kJ</em>

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

<em>efficiency of the Carnot's engine = 1 - </em>\frac{Tc}{Th}<em> </em>

eff. of the Carnot's engine = 1 - \frac{400}{600}

eff = 1 - 0.25 = 0.75

<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>

work done by heat engine = 2 kJ

<em>eff. of heat engine is gotten as = W/Q</em>

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>

6 0
3 years ago
A full-scale working model used to test a design solution by making actual observations and necessary adjustments.
zavuch27 [327]
The answer is prototype, or the first choice. hope this helps (:
3 0
3 years ago
A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at
Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, T_{s}=280^{o}C

Temperature of air, T_{\infinity}=20^{o}C

Velocity of air flow V=5.5m/s

The film temperature is determined as:

T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C

from the table, properties of air at 1 atm pressure

At T_{f}=150^{o}C

Thermal conductivity, K = 0.03443 W/m^oC; kinematic viscosity v=2.860 \times 10^{-5} m^2/s; Prandtl number Pr=0.70275

The reynolds number for the flow is determined as:

Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92

sice the obtained reynolds number is less than 2\times10^5, the flow is said to be laminar.

The nusselt number is determined from the relation given by:

Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}

Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11

The covective heat transfer coefficient is given by:

Nu_{cyl}=\frac{hD}{k}

Rewrite and solve for h

h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m

The rate of heat transfer from the wire to the air per meter length is Q=340.42W/m

6 0
3 years ago
Air is compressed steadily from 100kPa and 20oC to 1MPa by an adiabatic compressor. If the mass flow rate of the air is 1kg/s an
igomit [66]

Answer:

(a). 575 kJ/kg.

(b). 290kw.

Explanation:

We have the following set of information or parameters from the question above;

Pressure(1) = 100kPa, Pressure (2) = 1MPa, temperature(1) = b1= 12°C = 285K = 285kJ/kg, efficiency = 80% and the mass flow rate of the air = 1kg/s.

At a temperature of 12°C, we have the value from steam table; gx= 1.2, thus gx22 = 1.2 × (1000/100) = 12.

We have that the value for b12 = 517.

Therefore, the value for h2a can be calculated as;

80/100 = (517 - 285)/ (tp at exist) - 285.

0.8 = 232/ (tp at exist) - 285.

232 = 0.8 × (tp at exist) - 285).

232 = 0.8 (tp at exist) - 228 .

(tp at exist) = 575.

Therefore, the temperature 575 kJ/kg.

Thus, the required power input of the compressor = 1kg/s × ( 575 - 285) = 290kw.

6 0
3 years ago
A relatively nonvolatile hydrocarbon oil contains 4.0 mol % propane and is being stripped by direct superheated steam in a strip
hram777 [196]

Answer:

Number of Trays = Six (6)

Explanation:

Given that: y' = 25x' , in terms of molecular ratio, we can write it as

\frac{Y'}{1 + Y'} =25 \frac{X'}{1 + X'}  ......... 1

after plotting this we get equilibrium curve as shown in the attached picture.

inlet concentration and outlet concentration of liquid phase is

x₂ = 4% = 0.04 (inlet)

so that can be converted into molar

X_2 = \frac{x_2}{1-x_2} = \frac{0.04}{1-0.04} = 0.04167

and

x₁ = 0.2% = 0.002

X_1 = \frac{x_1}{1-x_1} = \frac{0.002}{1-0.002} = 2.004*10^{-3}

Now we have to use the balance equation a

\frac{G_s}{L_s} = \frac{X_2-X_1}{Y_2-Y_1} .............. a

here amount of solute is comparably lower than

Here we have

L = 300 kmol (total)

L_s = 300(1 - 0.04) = 288 kmol pure oil

G = G_s = 11.42 kmol

Y_1 = 0 , solvent free steam

substitute into the equation a

\frac{11.42}{288} = \frac{0.04167 - 2*10^{-3}}{Y_2 - 0}

Y₂ = 1.0003

Now plot the point A(X₁ , Y₁) and B(X₂ , Y₂) and join them to construct operating line AB.

Starting from point B, stretch horizontal line up to equilibrium curve and from there again go down to operating line as shown in the picture attached. This procedure give one count of tray and continue the same procedure up to end of operating.

at last count, the number of stage, gives 6.

∴ <em>Number of trays = 6</em>

5 0
4 years ago
Read 2 more answers
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