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JulsSmile [24]
2 years ago
7

108/49 In ? 108/48 Cd + ? In the equation above, what particle or type of radiation needs to be included on the right hand side

in order to balance it?
a. gamma
b. positron
c. beta
d. proton
e. alpha

Chemistry
1 answer:
Oksanka [162]2 years ago
7 0

Answer: b) positron

Explanation:

The antiparticle or the antimatter counterpart of the electron is called positron.

The positron has an electric charge of +1 e and same mass as an electron. When a positron collides with an electron, annihilation takes place.

Below is an attachnent that further explains this

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Answer: 100x

Explanation:

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8 0
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Read 2 more answers
At 650 K, the reaction MgCO3(s)⇌MgO(s)+CO2(g)MgCO3(s)⇌MgO(s)+CO2(g) has Kp=0.026Kp=0.026. A 10.0-L container at 650 K has 1.0 g
Ray Of Light [21]

Answer:

0.4076 g

Explanation:

Kp is the equilibrium constant based on pressure and depends only on gas substances. For a generic reaction

aA + bB ⇄ cC + dD

Kp = \frac{(pC)^c*(pD)^d}{(pA)^a*(pB)^b}, where pX is the pressure of X in equilibrium.

For the reaction Kp = pCO₂

pCO₂ = 0.026 atm

The system is in equilibrium at the beginning. The compression occurs at a constant temperature, so using Boyle's law

P1V1 = P2V2

0.026*10 = P2*0.1

P2 = 2.6 atm

The reaction will reach again the equilibrium, and pCO₂ = 0.026 atm, then the rest will form MgCO₃, which will be 2.6 - 0.026 = 2.574 atm.

By the ideal gas law:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature.

2.574*0.1 = n*0.082*650

53.3n = 0.2574

n = 4.83x10⁻³ mol

The stoichiometry of the reaction is 1 mol of MgCO₃ for 1 mol of CO₂, so it will form 4.83x10⁻³ mol of MgCO₃  .

The molar mass is:

MgCO₃: 24 g/mol of Mg + 12 g/mol of C + 3*16 g/mol of O = 84 g/mol

The mass formed is the molar mass multiplied by the number of moles:

m = 84x4.83x10⁻³

m = 0.4076 g

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