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2 years ago

8

Select the correct statement.

1 answer:

jok3333 [9.3K]2 years ago

7 0

D) basic ions contain some H+ ions

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According to the octet rule, an atom with two electron shells is most stable when it contains eight

Lorico [155]

According to the octet rule, an atom with two electron shells is most stable when it contains eight electrons. The octet rule is always based on valence electrons and their tendency to form bonds. <span />

4 0

4 years ago

WHAT BLOCK IN THE PICTURE HAS:

GREYUIT [131]

Answer:

the highest is "A" and the lower is "C"

7 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

How many grams are in 2.5 moles of PO4?

Inessa05 [86]

94.971361 grams would be the answer

8 0

4 years ago

How many acetone molecules are in a bottle of acetone with a volume of 445 mL ? (density of acetone = 0.788 g/cm3).

Mkey [24]

<h2>Hello!</h2>

The answer is: $3.63x10^{24}particles$

<h2>Why?</h2>

First, we need to calculate the molecular weight of the acetone (CH3COCH3)

So,

$C=12.011g/mol\\O=15.99g/mol\\H=1.008g/mol$

Then,

$CH3COCH3=12.011+(1.008)*3+12.011+15.999+12.011+(1.008)*3=58.08g/mol$

Second, we need to calculate the mass of the acetone in the bottle

We need to remember that $1cm^{3}=1mL$

So,

$445cm^{3}*\frac{0.788g}{cm^{3}}=350.66g$

Third, we need to calculate the number of moles:

$350.66g*\frac{1mol}{58.08g}=6.03moles$

Fourth, we need to calculate the number of atoms:

Remember, $1 mol=6.022x10^{23}particles$

Therefore,

$6.03moles*\frac{6.022x10^{23}particles }{1mol}=3.63x10^{24}particles$

Have a nice day!

4 0

4 years ago

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