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3 years ago

PLEASE HELP, NO BOTS PLEASE

Chemistry

2 answers:

ser-zykov [4K]3 years ago

7 0

Answer: Electrons= 7

Protons= 7

Neutrons= 7

Nitrogen has atomic number 7- the atomic number is the same as the number of protons and electrons in a neutral atom (an atom that isn’t an ion). To find the neutrons, we must subtract the atomic number or number of protons from the mass number which, in this case, is 14. 14-7 is 7 so this is the number of neutrons.

aleksandrvk [35]3 years ago

4 0

Protons: 7
electrons: 7
neutrons: 7

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ANDRES MEJIA-C...

BigorU [14]

Explanation:

Gases: more space therefore meaning that particles in the gas have the highest Kinetic energy therefore moving faster all around.

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3 years ago

Which has more energy, 500g of ice or 500g of steam

liraira [26]

Answer:

Steam because steam pushes out and ice stays dormant.

Explanation:

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3 years ago

Which of these are isotopes of each other? Boron-12, Carbon-12, Carbon-13, Nitrogen-13

Anika [276]

Carbon 13 I believe.
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4 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

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3 years ago

The measured dipole moment of a polar covalent compound is 1.11 D with a percent ionic character of 18.2%. What is the bond leng

TEA [102]

Answer:

80.2

Explanation:

we subtract 18.2 from the total percentage which is 100%

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3 years ago