A chemical process converts molten iron (III) oxide into molten iron and carbon dioxide by using a reducing agent of carbon monoxide. The process allows 10.08 kg of iron to be produced from every 16.00 kg of iron (III) oxide in an excess of carbon monoxide. Calculate the percentage yield of iron produced in this process.

Answers

Answer 1

Answer:

Answer:

percentage yield = 63%

Explanation:

The yield efficiency or percentage yield measure the amount of products that are formed from a given amount of reactant. For a percentage yield of 100, all the reactants are completely converted to product. Mathematically, the percentage yield is given by:

$percentage\ yield = (Actual\ yield)/(expected\ yield) * 100\nActual\ yield = 10.08kg\nExpected\ yield= 16.00kg\n\n\therefore percentage\ yield = (10.08)/(16.00) * 100 = 63 \%$

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A 78-percent efficient 12-hp pump is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s through a constant-diameter pipe. The free surface of the pool is 32 ft above that of the lake. Determine the irreversible head loss of the piping system, in ft, and the mechanical power used to overcome it. Take the density of water to be 62.4 lbm/ft3.

for a rankine cycle with one stage of reheat between turbines, there are how many relevant pressures?

Answers

The four relevant pressures in a Rankine cycle with one stage of reheat are P1, P2, P3, and P4.

For a Rankine cycle with one stage of reheat between turbines, there are typically four relevant pressures:

Boiler pressure (P1): This is the pressure at which the water is heated in the boiler before entering the first turbine.
High-pressure turbine outlet pressure (P2): This is the pressure at the outlet of the first turbine before the steam is sent to the reheater.
Reheat pressure (P3): This is the pressure at which the steam is reheated before entering the second turbine.
Low-pressure turbine outlet pressure (P4): This is the pressure at the outlet of the second turbine, which is also the condenser pressure.

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Air expands adiabatically through a nozzle from a negligible initial velocity to a final velocity of 300 m/s, what is the temperature drop of the air, if air is assumed to be an ideal gas for which CP = (7/2)R?

Answers

The temperature drop of air if air is assumed to be an ideal gas for which C_p = ⁷/₂R is; Δt = 1546 K

We are given;

Final velocity; v₂ = 300 m/s

C_p = ⁷/₂R

At constant pressure, the change in enthalpy is;

Δh = C_p × Δt

Now, from first law of thermodynamics;

h₂ + (v₂²/2) = h₁ + (v₁²/2)

We are told initial velocity is negligible and as such v₁ = 0 m/s

Thus;

h₂ + (v₂²/2) = h₁ + 0

(h₁ - h₂) = (v₂²/2)

Thus; Δh = v₂²/2

Finally;

C_p × Δt = v₂²/2

Δt = v₂²/2/(C_p)

Δt = (300²/2)/(⁷/₂R)

where R is ideal gas constant = 8.314 Kj/kg.mol

Thus;

Δt = (300²/2)/(⁷/₂ × 8.314)

Δt = 1546 K

Read more at; brainly.com/question/24188841

Answer:

ΔH+U²/2=0

and

ΔH= $C{p$ ×ΔT

∴to get the temperature drop of air, you make ΔT subject of the formula

ΔT=-U²/2Cp

=-300²/2× $(7)/(2)$ ×8.314

∴ΔT=-1546K

Explanation:

Which of the following was an effect of world war 2 on agricultural industry

Answers

Answer:

Option C..Farmers saught new technology to help with the workload

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How much heat is lost through a 3’× 5' single-pane window with a storm that is exposed to a 60°F temperature differential?A. 450 Btu/hB. 900 Btu/hC. 1350 Btu/hD. 1800 Btu/h

Answers

Answer:

A. 450 btu/h

Explanation:

We solve this problem by using this formula:

Q = U x TD x area

U = U value of used material

TD = Temperature difference = 60°

Q = heat loss

Area = 3x5 = 15

We first find U

R = 1/u

2 = 1/U

U = 1/2 = 0.5

Then when we put these values into the formula above, we would have:

Q = 0.5 x 15 x 60

Q = 450Btu/h

Therefore 450btu/h is the answer

The specific volume of mercury is .00007 m^3/Kg. What is its density in lbm/ft^3?

Answers

Answer:

891.027 lbm/ft³

Explanation:

See it in the pic

Consider the binary eutectic copper-silver phase diagram in Fig. P8.22. Make phase analyses of an 88 wt % Ag−12 wt % Cu alloy at the temperatures (a) 1000°C, (b) 800°C, (c) 780°C + ΔT, and (d) 780°C − ΔT. In the phase analyses, include: (i) The phases present (ii) The chemical compositions of the phases (iii) The amounts of each phase (iv) Sketch the microstructure

Answers

Answer:

answer is 0 ok answer is 0

Explanation:

Other Questions

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

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