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Serjik [45]
3 years ago
13

Which form of emission is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass

numbers?
a. alpha particle
b. beta particle
c. gamma ray
Chemistry
2 answers:
lora16 [44]3 years ago
8 0
<span>The form of emission that is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass numbers is the gamma ray. 

Gamma rays have no electrical charge nor atomic mass. It is only used to balance in a nuclear reaction, but gamma ray is considered powerful.

So, letter C. is the answer.
Hope this helps!</span>
schepotkina [342]3 years ago
4 0

Answer: c. gamma ray

Explanation: There are various process in which a radioactive nuclei decays:

1) Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units.

_A^Z\textrm{X}\rightarrow _{A-2}^{Z-4}+_2^4\alpha

A= mass number

Z= atomic number

2) Beta particle: It is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. This is also known as \beta ^+-decay. In this the mass number remains same.

_A^Z\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e

3) Gamma ray emission: It is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. This decay releases \gamma -radiations. This process does not change the mass number.

_A^Z\textrm{X}^*\rightarrow _A^Z\textrm{X}+_0^0\gamma

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3 years ago
At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5
Annette [7]

Answer:

4600s

Explanation:

2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}

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-\frac{d[B]}{dt}=k[B] - - -  -\frac{d[B]}{[B]}=k*dt

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.  

-\frac{d[P(B)]}{P(B)}=k*dt  

-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt

Integrating we get:

\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt

-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})

Clearing for t2:

\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}

ln[P(N_{2}O_{5})]=ln(650)=6.4769

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4 0
3 years ago
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frutty [35]

Answer:

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The first empirical study on the behaviour of a mixture of gases was carried out by John Dalton. He established the effects of mixing gases at different pressures in the same vessel.

Dalton's law states that,the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture of gases. When a gas is collected over water, the gas also contains some water vapour. The partial pressure of the gas will now be given as; total pressure of gas mixture - saturated vapour pressure of water (SVP) at that temperature.

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Answer:

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