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3 years ago

Write a nuclear equation to represent radium-226 decaying to radon-222 plus an alpha particle.

Chemistry

1 answer:

valentina_108 [34]3 years ago

4 0

Answer:

22688Ra→l22286Rn+l42He

Explanation:

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation:

22688Ra→l22286Rn+l42He

Note that the sum of the subscripts (atomic numbers or charges) is the same on each side of the equation.

Also, the sum of the superscripts (masses) is the same on each side of the equation.

Now try figuring out Am-241

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3 years ago

A chemist prepares a solution of magnesium chloride MgCl2 by measuring out 49.mg of MgCl2 into a 100.mL volumetric flask and fil

Flura [38]

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Explanation:

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Formula used :

$Molarity=\frac{n\times 1000}{V_s}$

where,

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Now put all the given values in the formula of molarity, we get

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Therefore, the molarity of solution will be $5.2\times 10^{-3}mole/L$

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As 1 mole of $MgCl_2$ gives 2 moles of $Cl^-$

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3 years ago

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2. ENERGY LEVELS!!!

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