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3 years ago

8

Engineers often attempt to capture and reuse energy or to use renewable resources to minimize harm in the environment. This is R

eferred to as ___

1 answer:

bogdanovich [222]3 years ago

3 0

I believe that this is reffered to as Green Design.

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Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

natka813 [3]

Answer:

Explanation:

You need to remember that the oxidation number of H is +1, except when it is in a metal hydrites like NaH, where its oxidation number is -1. Then, the oxidation number of O is -2, but in peroxides is -1. So with these rules you just have to multiply the ox. number with the name of atoms and all the elements in the reaction must sum 0.

5 0

3 years ago

Text 8.7 Using the virial equation of state for hydrogen at 298 K given in problem 7 (text 8.6), calculate a. The fugacity of hy

gregori [183]

Answer:

a). P = 688 atm

b). P = 1083.04 atm

c).Δ G = 16.188 J/mol

Explanation:

a). Fugacity 'f' can be calculated from the following equations :

$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$

where, P = pressure , Z = compressibility

Now, the virial equation is :

$PV=R(1+6.4\times 10^{-4} P)$ ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

$Z=1+6.4\times 10^{-4} P$

So from the fugacity equation ,

$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

$f=Pe^{6.4 \times 10^{-4} P}$

Putting the value of P = 500 atm in the above equation, we get,

f = 688 atm

b). Given f = 2P

$2P=PE^{6.4 \times 10^{-4}P}$

$2=E^{6.4 \times 10^{-4}P}$

$\ln 2 = 6.4 \times 10^{-4}P$

∴ P = 1083.04 atm

c). dG = Vdp -S dt at constant temperature, dT = 0

Therefore, dG = V dp

$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp$

$=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$

$=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$

$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$

Δ G = 16.188 J/mol

7 0

3 years ago

Write the uses of all elements which are given in the periodic table

Karolina [17]

Answer:

write the uses of all elements which are given in the periodic table

6 0

3 years ago

Which states of matter have particles that move independently of one another with very little attraction?

bonufazy [111]

Answer:

Gas

Explanation:

The gaseous state has very loose and unorganized structuring of particles, making them have little attraction and move independently.

4 0

3 years ago

How do you know sodium is highly reactive?

Stells [14]

<span>Na (sodium) is highly electropositive. Its has 1 electron in its outermost orbit which is transferred to an electronegative atom to form an ionic bond.

It only needs to get rid of one valence electron to take part in a reaction. That's how it's highly reactive.</span>

4 0

3 years ago