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3 years ago

6

In ionic bonds what happens to electrons

2 answers:

Nana76 [90]3 years ago

5 0

Answer:

C: metals donate electrons to non metals

Explanation:

bezimeni [28]3 years ago

3 0

Answer : The correct option is, (C) metals donate electrons to non metals

Explanation :

Covalent compound : Covalent compounds are the compound in which the atoms are covalently bonded. The covalent bonds are formed by the equal sharing of the electrons and these bonds are formed between two non-metals.

Ionic compound : Ionic compounds are the compound in which atoms are bonded through the ionic bond. The ionic bonds are formed by the complete transfer of electrons. That means the bond is formed when the metals donate electrons to the non-metals.

These bonds are formed between one metal and one non-metal.

Hence, the correct option is, (C) metals donate electrons to non metals

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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

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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 ,

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

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

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

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

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f = 688 atm

b). Given f = 2P

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

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

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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

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