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2 years ago

PLEASE ANSWER NEED BY MIDNIGHT TONIGHT!!!!!!

Chemistry

1 answer:

maksim [4K]2 years ago

5 0

The equation for carbon-14 emission by Radium-223 nuclei is given below:

$^{223}_{88}Ra \rightarrow\: ^{209}_{82}Pb + \:^{14}_{6}C$

<h3>What is radioactivity?</h3>

Radioactivity is the spontaneous decay of a substance with emission of radiation.

The equation for carbon-14 emission by Radium-223 nuclei is given below:

$^{223}_{88}Ra \rightarrow\: ^{209}_{82}Pb + \:^{14}_{6}C$

In conclusion, the emission of carbon-14 by Radium-223 nuclei produces Lead-209 nuclei.

Learn more about radioactivity at: brainly.com/question/3603596

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When cyclopentane undergo free-radical substitution with bromine (Br2 /Heat) the product:

Artemon [7]

Explanation:

b. Bromo cyclopentane + HBr

4 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

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3 years ago

The burning of gasoline in an automobile engine is an example of a(n) _____.

inysia [295]

It is an exothermic reaction because the heat is released.

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3 years ago

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iris [78.8K]

Answer:

The answer to your question is given below

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + 2HCl —> ZnCl2 + H2

Thus, we can write out the atoms present in both the reactant and the product by doing a simple head count. The atoms present are listed below:

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Zn >>>>>>>> 1 >>>>>>>>>> 1

H >>>>>>>>> 2 >>>>>>>>> 2

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3 years ago

A measure of how closely individual measurements agree with one another 1 point<br> is

rewona [7]

Precision

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3 years ago