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3 years ago

Select the correct answer from each drop-down menu.

Physics

1 answer:

bagirrra123 [75]3 years ago

8 0

Answer:

(1) An object that’s negatively charged has more electrons than protons.

(2) An object that’s positively charged has fewer electrons than protons.

(3) An object that’s not charged has the same number of electrons than protons.

Explanation :

Objects have three subatomic particles that are Electrons, protons, and neutrons.

Protons and neutrons are found in the nucleus and electrons rotate or move outside the nucleus. Naturally, protons are positively charged, neutrons have no charge, and electrons are negatively charged.

Therefore, an object that is negatively charged has more electrons than protons. An object that is not charged has the same number of electrons than protons. An object that is positively charged has fewer electrons than protons.

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ruslelena [56]

Answer: a) $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n$

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Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n$

Now, as the mass on both reactant and product side must be equal:

$235+1=87+146+x$

$x=3$

Thus three neutrons are produced and nuclear equation will be: $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n$

b) For the another fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

A = 140

To calculate Z:

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

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}$

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