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kondaur [170]
1 year ago
8

A decay series starts with the synthetic isotope ²³⁹₉₂U. The first four steps are emissions of a β⁻ particle, another β⁻, an a p

article, and another α. Write a balanced nuclear equation for each step. Which natural series could start by this sequence?
Chemistry
1 answer:
VladimirAG [237]1 year ago
7 0

<u>Thorium series</u> could start by this sequence.

<h3>Brief explanation</h3>

To write balanced equations for nuclear decay processes. It's important to remember that the mass number and the atomic numbers must be balanced. And so what that means is that if we look at an elements nuclear symbol, the atomic number is the bottom number and the top number, the superscript, is the mass number, and so when we add them up on both sides, they have to be equal. There are two different ways in which decay can occur.

In this, series one is through beta decay, which means that the following particle is produced. The other is Alpha Decay, which produces this particle. Both are products. So if we start off with uranium to 39 you read it in nuclear notation, which means we have to find the atomic number just 92 and it undergoes beta decay.

So that means that it produces this particle find the second particle we used the atomic number, so 92 equals minus one plus x, where X equals 93 which is Neptune IAM. The mass number of our new isotope is zero plus X equals to 39 where X equals to 39. This product becomes the reactant in my next decay, which is also a beta decay. And to find the unknown element we do the same here.

Except for that it's 93 equals minus one plus x, where X is 94 which is P u plutonium, and the mass number is zero plus X equals to 39 or to 39. The next decay starts with the isotope that we just form to 39 p. U. This time it's an Alpha decay. So we produce this particle to find the unknown. Element 94 equals two plus x, where X equals 92 which takes us back to uranium.

Find the mass number of this isotope 2 39 equals four plus X, where X equals to 35. Finally, for the last decay, you have another Alpha decay starting with uranium to 35 making an alpha particle. The atomic number will be 90 which is T H and the top is 2 31 For the mass number. This begins the natural decay, series of thorium .

Learn more about chemical decay

brainly.com/question/1898040

#SPJ4

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The violet light is the answer!

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The amount of energy it takes to remove an electron from an atom is
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3 years ago
PLEASE HELP ASAP WILL GIVE BRAINLIEST
kirill115 [55]

The amount of heat required to convert H₂O to steam is :  382.62 kJ

<u>Given data :</u>

Mass of liquid water  ( m ) = 150 g

Temperature of liquid water = 43.5°C

Temperature of steam = 130°C

<h3 /><h3>Determine the amount of heat required </h3>

The amount of heat required = ∑ q1 + q2 + q3 ----- ( 1 )

where ;

q1 = heat required to change Temperature of water from 43.5°C to 100°C .  q2 = heat required to change liquid water at 100°C to steam at 100°C

q3 = heat required to change temperature of steam at 100°C to 130°C

  • For q1

M* S_{water}*ΔT

= 150 * 4.18 * ( 100 - 43.5 )

= 35425.5 J

  • For q2

moles * ΔHvap

= (150 / 18 )* 40.67 * 1000

=  338916.67 J

  • For q3

M * S_{steam} * ΔT

= 150 * 1.84 * ( 130 -100 )

= 8280 J

Back to equation ( 1 )

Amount of heat required = 35425.5  + 338916.67 + 8280 = 382622.17 J

                                                                                               ≈ 382.62 kJ

Hence we can conclude that The amount of heat required to convert H₂O to steam is :  382.62 kJ.

Learn more about Specific heat of water : brainly.com/question/16559442

6 0
3 years ago
ubstance A is a nonpolar liquid and has only dispersion forces among its constituent particles. Substance B is also a nonpolar l
LenaWriter [7]

Answer:

Their particles exhibit the same type of intermolecular interaction

Explanation:

In chemistry, we commonly say that 'like dissolves like'. This implies that polar solvents dissolves polar solutes while nonpolar solvents dissolve nonpolar solutes.

This phenomenon of 'like dissolves like' is possible because, the dissolution of one substance in another involves intermolecular interaction between the solute and solvent molecules.

If the molecules of solute and solvent are both nonpolar and have about the same magnitude of intermolecular (dispersion) forces, interaction between the both molecules is significant hence the solute dissolves completely in the solvent.

6 0
3 years ago
How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add
Nostrana [21]

Answer :

The correct answer   for Mass of Na₂HPO₄ = 4.457 g and mass of  NaH₂PO₄  = 8.23 g

Given :  pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of  Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of  Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from  NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

<u>Step 1 : To find pka </u>

H₂PO₄⁻  <=> HPO₄²⁻  

The above reaction has pka = 7.2 ( from image shown )

<u>Step 2 : Plug values in Hasselbalch- Henderson equation </u>.

Hasselbalch -Henderson equation is to find pH  for buffer solution which is as follows :

pH = pka + log\frac{[A^-]}{[HA]}

pH = 6.86         pKa = 7.2

6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Subtracting  both side by 7.2

6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

-0.34 =  log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Removing log

10^-^0^.^3^4 =   \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457 ---------------- equation (1)

<u>Step 3 : To find  molarity of H₂PO₄⁻ and HPO₄²⁻</u>

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence,  [H₂PO₄⁻ ] + [ HPO₄²⁻ ] =  0.1 M

Assume [H₂PO₄⁻ ] = x

So ,  [x ] + [ HPO₄²⁻ ] =  0.1 M

[ HPO₄²⁻ ] =  0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ]

[H₂PO₄⁻ ]  = x

 [ HPO₄²⁻ ] = 0.1 - x

Equation (1) = >\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457

Plug value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ] ( from step 3 ) into equation (1)  as :

\frac{[0.1 - x ]}{[ x]} = 0.457

Cross multiplying

0.1 - x  = 0.457 x

Adding x on both side

0.1 -x + x = 0.457 x + x

0.1  = 1.457 x

Dividing both side by 1.457

\frac{0.1}{1.457} = \frac{1.457 x }{1.457}

x = 0.0686 M

Hence , [H₂PO₄⁻ ]  = x  = 0.0686 M

 [ HPO₄²⁻ ] = 0.1 - x

 [ HPO₄²⁻ ]  =   0.1 - 0.0686  

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of  H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M  or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M  or 0.0314 mole per 1 L

Since  that volume of buffer solution  is 1 L , so Molarity  = mole

Hence Mole of NaH₂PO₄  = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

<u>Step 6 : To find mass  of Na₂HPO₄  and NaH₂PO₄ </u>

Moles of  Na₂HPO₄  and NaH₂PO₄  can be converted to their masses using molar mass as follows :

Molar mass of  Na₂HPO₄  = 141.96 \frac{g}{mol}

Molar mass of NaH₂PO₄ = 119.98 \frac{g}{mol}

Mass (g) = mole (mol)* molar mass(\frac{g}{mol})

Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}

Mass of Na₂HPO₄ = 4.457 g

Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}

Mass of  NaH₂PO₄  = 8.23 g

5 0
3 years ago
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