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3 years ago

Two protons and two neutrons are released as a result of this reaction.

Chemistry

2 answers:

Ierofanga [76]3 years ago

6 0

Answer:

The answer is A

Explanation:

Murljashka [212]3 years ago

3 0

Answer: The particle released in the given reaction is one alpha particle

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{222}_{86}\textrm{Rn}\rightarrow ^A_Z\textrm{X}+^{218}_{84}\textrm{Po}$

To calculate A:

Total mass on reactant side = total mass on product side

222 = A + 218

A = 4

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

86 = Z + 84

Z = 2

The isotopic symbol of unknown element is $_{2}^{4}\textrm{He}$ . Another name of helium atom is alpha particle.

Hence, the particle released in the given reaction is one alpha particle

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A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added

Ulleksa [173]

Answer: The pH change of the buffer is 0.30

Explanation:

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})$ .....(1)

We are given:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.306M$

$[C_6H_5NH_2]=0.418M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{initial}=14-8.99=5.01$

To calculate the molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-$

Initial: 0.418 0.124 0.306

Final: 0.294 - 0.430

Calculating the pOH by using using equation 1:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.430M$

$[C_6H_5NH_2]=0.294M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{final}=14-9.29=4.71$

Calculating the pH change of the solution:

$\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30$

Hence, the pH change of the buffer is 0.30

8 0

3 years ago

How many moles of boron (B, 10.81 g/mol) are in 37.8 grams of boron? [?] moles

madreJ [45]

Answer:

37.8g/ 10.81g/mol = 3.4968...moles

3 0

2 years ago

How does water’s ability to form hydrogen bonds explain why ice is less dense than liquid water?

Karolina [17]

Answer:

As water freezes, a crystalline structure preserved by hydrogen bonding is formed by water molecules. Less dense than liquid water is solid water, or ice. Ice is less dense than water since molecules are pulled farther apart by the direction of hydrogen bonds, which decreases density.

Explanation:

3 0

3 years ago

The ro– group is the nucleophile. think about which group in the alkyl halide (iodoethane) is the leaving group. keep in mind th

jonny [76]

When Nucleophile (Alkoxide) reacts with Alkyl Halide (Substrate), Ether (product) is formed with the elimination of Leaving group (Iodide in this case).

This is SN² a nucleophilic substitution reaction which takes place in primary alkyl halides. Reaction is given as,

8 0

3 years ago

How are catapults an example of a third class lever? Please explain!

Katen [24]

I hope that you can understand this!!! Lol

The thing that you have to pull back to release with, that would be considered a third class lever.

I hope this helps. :)

3 0

3 years ago