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mario62 [17]
3 years ago
10

What is the daughter nucleus (nuclide) produced when 227th undergoes alpha decay?

Chemistry
1 answer:
Elina [12.6K]3 years ago
6 0

Answer: -

²²³Ra is the daughter nuclide produced when ²²⁷Th undergoes alpha decay

Explanation: -

When alpha decay occurs, the mass number of the parent decreases by 4 and the atomic number decreases by 2.

Mass number of ²²⁷Th =227

Atomic number of ²²⁷Th = 90

Mass number of daughter = 227 - 4 = 223

Atomic number of daughter = 90 - 2 = 88

88 is the atomic number of Ra Radium.

Thus the daughter is ²²³Ra

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The atomic symbol Superscript 206 subscript 82 upper P b. represents lead-206 (Pb-206), an isotope that has 82 protons and 124 n
Serggg [28]

Answer:

\left \{ {{y=206} \atop {x=82}}Pb \right.

Explanation:

isotopes are various forms of same elements with different atomic number but different mass number.

Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are

  • Alpha particle emission  \left \{ {{y=4} \atop {x=2}}He \right.
  • Beta particle emission    \left \{ {{y=0} \atop {x=-1}}e \right.
  • gamma radiation             \left \{ {{y=0} \atop {x=0}}γ \right.

in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.

Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below

\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right +  \left \{ {{y=0} \atop {x=0}}γ\right.

Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right

8 0
2 years ago
Read 2 more answers
Can a bad thermostat cause radiator to blow
erastova [34]
I don't exactly know but probably so.
3 0
3 years ago
A silver necklace has a density of 10.5 g/mL and a volume of 1.5 mL. What is the mass of the silver necklace?
melamori03 [73]

Answer:

15.75g

Explanation:

Rearrange the formula: density = mass/volume, so that mass is the subject: mass = density × volume

Now substitute (plug in) the known values into the formula:

mass = 10.5 × 1.5 ⇒ mass = 15.75g

The mass of the silver necklace is 15.75g

Hope this helps!

4 0
2 years ago
A solution of H 2 SO 4 ( aq ) H2SO4(aq) with a molal concentration of 8.01 m 8.01 m has a density of 1.354 g / mL . 1.354 g/mL.
anzhelika [568]

Answer:

[H₂SO₄] = 6.07 M

Explanation:

Analyse the data given

8.01 m → 8.01 moles of solute in 1kg of solvent.

1.354 g/mL → Solution density

We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g

Mass of solvent = 1kg = 1000 g

Mass of solution = 1000g + 785.4 g = 1785.4 g

We apply density to determine the volume of solution

Density = Mass / volume → Volume = mass / density

1785.4 g / 1.354 g/mL = 1318.6 mL

We need this volume in L, in order to reach molarity:

1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L

Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M

4 0
2 years ago
Read 2 more answers
How are the following aspects of a reaction affected by the addition of a catalyst? 1) activation energy of the reverse reaction
snow_tiger [21]

These are four questions and four answers.

Answers:

1) activation energy of the reverse reaction

     b. Decreased

2) Rate of the forward reaction

    a. Increased

3) Rate of the reverse reaction

    a. Increased

4) Activation energy of the forward reaction

    b. decreased

Explanation:

<em>Activarion energy</em> is the energy required by the reactants to form the intermediate transition state and become products.

<em>Catalysts</em> are substances that change the path of the chemical reactions, lowering the activation energy, and thus speeding up the rate of the reactions, since the products can reach the new lower activation energy faster.

The equilibrium reactions are the chemical process in which two reactions, the <em>forward and the reverese reactions</em>, occur simultaneously and at the same rate.  The equlibrium reactions may be represented by:

  • A ⇄ B

Where A → B is the direct or forward reaction, and A ← B is the reverse reaction (note the inversed arrow, from right to left).

For the direct reaction A represents the reactants and B represents the products. On the other hand, B represents the reactants and A represents the reactants of the reverse reaction and A. This, is A is the reactant of the forward reaction and the product of the reverse reaction, while B is the reactant of the reverse reaction and the product of the forward reaction.

Since, <em>the addition of a catalyst</em> lowers the activation energy of the process, the new activation energy is lower for both the forward and the reverse reaction, meaning that:

1. <em>The activation energy of the reverse reaction is decreased</em> (option b. of the first question)

2.<em> The rate of the forward reaction is increased</em> (option a. of the second question)

3. <em>The rate of the reverse reaction is increased</em> (option a. of the third question).

4. <em>Activation energy of the forward reaction is decreased</em> (option b. of the fourth question).

In summary, the addition of a catalyst decreases the activation energy for both forward and reverse reactions, and increases the rate of both forward and reverse reactions.

3 0
3 years ago
Read 2 more answers
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