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3 years ago

The equations are not balanced. Which equation would have the same coefficients in the same order as 2CO2 + 3H20 → C2H6O + 3O2?

Physics

2 answers:

mr Goodwill [35]3 years ago

6 0

Answer: A. $2 Al+3FeO\rightarrow Al_2O_3+3 Fe$

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

A. $2 Al+3FeO\rightarrow Al_2O_3+3 Fe$

B. $CH_4+2O_2\rightarrow CO_2+2H_2O$

C. $PCl_5+4H_2O \rightarrow 5HCl +H_3PO_4$

D. $FeCl_3+3NH_4OH\rightarrow Fe(OH)_3+3NH_4Cl$

Thus the coefficients of 2, 3, 1 and 3 as present in $2CO_2+3H2O\rightarrow C_2H_6O+3O_2$ will be same in $2 Al+3FeO\rightarrow Al_2O_3+3 Fe$ .

KiRa [710]3 years ago

5 0

<span>A: Al + FeO → Al2O3 + Fe

Hope it helps!
</span>

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3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

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p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

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3 years ago

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igor_vitrenko [27]

Answer:

0.231 N

Explanation:

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$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

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