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ElenaW [278]
2 years ago
15

Which list of radioisotopes contains an alpha emitter, a beta emitter, and a positron emitter?

Chemistry
2 answers:
MArishka [77]2 years ago
6 0

Answer: The correct option is 3.

Explanation: Radioisotopes which emits alpha-particle are known as alpha-emitters. These radioisotopes undergo alpha-decay.

The radioisotopes which emits beta-particle (_{-1}^0\beta ) are known as beta-emitters. These radioisotopes undergo beta-minus decay. In this decay a neutron gets converted to a proton and an electron.

The radioisotopes which emits positron-particle (_{+1}^0\beta ) are known as positron-emitters. These radioisotopes undergo beta-plus decay. In this type of decay a proton gets converted to a neutron.

From the given options,

Option 1: All the three radioisotopes undergoes beta-minus decay.

Option 2: Cs-137 and Tc-99 radioisotopes undergo beta-minus decay.

Fr-220 is a radioisotope which undergoes alpha-decay.

Option 3: Radioisotope Kr-85 undergoes beta-minus decay.

_{36}^{85}\textrm{Kr}\rightarrow _{37}^{85}\textrm{Rb}+_{-1}^0\beta

Radioisotope Ne-19 undergoes positron decay.

_{10}^{19}\textrm{Ne}\rightarrow _{9}^{19}\textrm{F}+_{+1}^0\beta

Radioisotope Rn-222 undergoes alpha decay.

_{86}^{222}\textrm{Rn}\rightarrow _{84}^{218}\textrm{Po}+_{2}^4\alpha

Option 4: All the three radioisotopes undergoes beta-minus decay processes.

Hence, from the above information, the correct option is 3.

vodka [1.7K]2 years ago
4 0
The answer is (3), they are β-, β+ and α decay mode. For (1), they are β-, β- and β- decay mode. For (2), they are β-, α and β-. For (4), they are α, α and α decay mode.
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Determine the identity of each element based upon the following data.
svet-max [94.6K]

Ionization energy is defined as the energy required to remove the electron requiring the least amount of energy in the atom.

When valence electrons are being removed, the ionization energy will be quite low. However, once there is a spike in the number (In this case, IE6 starts at 8496 and jumps to 27110 in IE7), that indicates that a core electron is being removed.

Since there are 6 ionization energies that are relatively low before the spike in energy in IE(7), it can be inferred that this period 3 element has 6 valence electrons.

Because of this, we simply have to go from left to right in period 3 six times. (Make sure to skip over the transition metals if you're counting past period 3, since their valence electrons vary)

Sulfur is the mystery element

Good luck! If you have any questions, just ask! :))

3 0
2 years ago
Witch of the following isotopes would most likely be unstable and therefore radioactive?
STALIN [3.7K]
The answer is D; Mercury-194
All of the others are not when I looked them up

5 0
2 years ago
Can 750 mL of water dissolve 0.60 mol of gold(III) chloride (AuCl3)?
nydimaria [60]

Answer:

Yes  

Explanation:

1. Mass of 0.60 mol of AuCl₃  

\text{Mass} = \text{0.60 mol} \times \dfrac{\text{303.33 g}}{\text{1 mol}} = \text{184 g}

2. Mass of AuCl₃ in 750 mL

The solubility of AuCl₃ is 68 g/100 mL.

In 750 mL of water, you can dissolve

\text{Mass of AuCl}_{3} = \text{750 mL} \times \dfrac{\text{68 g}}{\text{100 mL}} = \text{510 g AlCl}_{3}

∴ Yes, 750 mL of water can dissolve 0.60 mol of AuCl₃.

3 0
2 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}

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

-\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

ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333

t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s

4 0
2 years ago
Which of the following actions cannot induce voltage in a wire?
MariettaO [177]
Your answer is D. Since there is little to no magnetic field to  wire, if it is copper which most wires are, there will be no voltage in a wire.
7 0
2 years ago
Read 2 more answers
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