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3 years ago

Which represents the balanced nuclear equation for the beta plus decay of C-11?

Chemistry

2 answers:

Nookie1986 [14]3 years ago

8 0

Answer:

Superscript 11 subscript 6 upper C right arrow superscript 11 subscript 5 upper B plus superscript 0 subscript plus 1 e.

Explanation:

Beta decay:

Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle. In beta positive decay proton is converted into neutron and positron, thus the atomic number is decreased by one.

¹¹₆C → ¹¹₅B + ⁰₊₁e

In beta minus decay the one electron is ejected and neutron is converted into proton.

¹⁴₆C → ¹⁴₇N + ⁰₋₁e

Properties of beta radiations:

The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminium is used to block the beta radiation .

Kobotan [32]3 years ago

6 0

The answer is A 100%. I just took the test.

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vagabundo [1.1K]

It is more slippery, and it is heavier

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3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

Describe the PURPOSE of Moh's hardness scale. (What information does it tell you)

Lerok [7]

Answer: The Mohs Hardness Scale is used as a convenient way to help identify minerals. A mineral's hardness is a measure of its relative resistance to scratching, measured by scratching the mineral against another substance of known hardness on the Mohs Hardness Scale.

Explanation: Hope this helped! :)

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3 years ago

Which of the following is an inference?

Len [333]

Answer:

the answer is d

Explanation:

but wusup shawty

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3 years ago

All reversible reactions will reach chemical _____.

xeze [42]

D.) <span>All reversible reactions will reach chemical "Equilibrium"

Hope this helps!</span>

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3 years ago