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3 years ago

What is emitted when the radioactive nucleus of 131 53i decays to form the stable isotope of xenon 131 54xe?

1 answer:

3 0

It is a beta particle that is being emitted when the radioactive nucleus of 131 / 53 I decays to form the stable isotope of xenon 131 / 54 Xe. It is written as 0/-1 e-. It is a high speed and high energy positron or electron during beta decay. An unstable nucleus containing an excess of neutrons would unergo a beta decay wherein the neutron is transformed into a proton, and electron and an antiparticle. The electrons emitted are not the electrons from the shells but are being produced when a neutron splits firming a proton and an electron. These particles would be able to pass in paper but can be blocked by an aluminum foil.

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Plzz help me out guys ......10 points rewarded.....!!!!!

GaryK [48]

number 5 is 1 : 1

number1 is also 1:1 ibelieve but it could be 2:1 just like number 5 but im postive its 1:1 for nummer 1 and number 5

4 0

3 years ago

Why can’t an atom lose or gain a proton

Veronika [31]

Explanation:

Atoms never gain protons; they become positively charge only by losing electrons. A positive ion is called a cation (pronounced: CAT-eye-on). You may have notice that the number of neutrons in each of these ions was not specified.

6 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

6 0

3 years ago

Which aldehyde is an intermediate in the reduction of ethyl benzoate with lithium aluminum hydride?

garri49 [273]

Answer:

tetrahedral aldehyde

Explanation:

The reaction begins with a hydride nucleophile reacting with the ester carbonyl carbon to form the tetrahedral intermediate.

The carbonyl reforms to produce an aldehyde with the loss of the alkoxide ion.
The resulting aldehyde undergoes a subsequent reaction with a hydride nucleophile to form another tetrahedral intermediate. The carbonyl is not able to reform, because there are no stable leaving groups.
Therefore, the tetrahedral intermediate is protonated to produce a primary alcohol.

6 0

3 years ago

What would be the bond order for He 2 2- molecule

LuckyWell [14K]

Answer:

In He2 molecule,

Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.

Molecular Orbitals thus formed are:€1s2€*1s2

It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .

Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2

Bond Order =Nb-Na/2

Bond Order =2-2/2=0

Since the bond order is zero so that He2 molecule does not exist.

Explanation:

8 0

3 years ago