How do I balance a chemical reaction like one of these NaBr + Ca(OH)2 ➡️ CaBr2 + NaOH

Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation

nikdorinn [45]

Answer: The isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

Explanation:

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

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

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

A = 139

To calculate Z:

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

92 + 0 = Z + 36 + 0

Z = 56

The isotopic symbol of barium is $_{56}^{139}\textrm{Ba}$

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

To calculate Z:

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

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is $_{38}^{89}\textrm{Sr}$

Hence, the isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

7 0

3 years ago

How many moles of 0.225 M CaOH2 are present in 0.350 L of solution?

weeeeeb [17]

Answer : The number of moles of solute $Ca(OH)_2$ is, 0.0788 moles.

Explanation : Given,

Molarity = 0.225 M

Volume of solution = 0.350 L

Formula used:

$\text{Molarity}=\frac{\text{Moles of }Ca(OH)_2}{\text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$0.225M=\frac{\text{Moles of }Ca(OH)_2}{0.350L}$

$\text{Moles of }Ca(OH)_2=0.0788mol$

Therefore, the number of moles of solute $Ca(OH)_2$ is, 0.0788 moles.

7 0

3 years ago

Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:

Maru [420]

Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and $1s^22s^22p^6$

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1$

This element will easily loose 1 electron and form $Rb^+$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

$1s^22s^22p^3$

This element will easily gain 3 electrons and form $N^{3-}$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

$1s^22s^22p^6$

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^5$

This element will easily gain 1 electron and form $Br^-$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

4 0

3 years ago

If I have 2 mol NaOH in 3 L of solution, what is the molarity of my solution?

qwelly [4]

Answer:

0.67 mole/litre

Explanation:

the molarity equall no. of moles ÷ volume of sol.

3 0

3 years ago

PLZ HELP! Give me a really (winner) good idea for a kids science fair! My little sister has to do one through my school and I ne

OverLord2011 [107]

Explanation:amiga

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extrañar

8 0

3 years ago