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ElenaW [278]
4 years ago
15

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

Chemistry
2 answers:
MArishka [77]4 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]4 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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Complete combustion of a 0.350 g sample of a compound in a bomb calorimeter releases 14.0 kJ of heat. The bomb calorimeter has a
dexar [7]

Answer:

Its final temperature is 25.8 °C

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal-Tinitial)

When a body transmits heat there is another that receives it. This is the principle of the calorimeter. Then the heat released by the compound will be equal to the heat obtained by the calorimeter.

In this case, you know:

  • Q= 14 kJ= 14,000 J
  • c= 3.55  \frac{J}{g*C}
  • m=1.20 kg= 1200 g (1 kg=1000 g)
  • Tfinal= ?
  • Tinitial= 22.5 °C

Replacing:

14,000 J= 3.55 \frac{J}{g*C}*1200 g*(Tfinal-22.5C)

Solving:

\frac{14,000J}{3.55\frac{J}{g*C} *1200 g} =T final - 22.5C

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<u><em>Its final temperature is 25.8 °C</em></u>

3 0
3 years ago
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3 Apply Use context clues to write your own definitions for the words definite and occupy.
motikmotik

Answer:

Explanation:

own definitions for the words definite and occupy.

Example sentence

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definite:

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3 0
2 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
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Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
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Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
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Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

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