All categories

3 years ago

What is a decay chain

2 answers:

7 0

In nuclear science, the decay chain refers to a series of radioactivedecays of different radioactive decay products as a sequential series of transformations. It is also known as a "radioactive cascade".

ehidna [41]3 years ago

3 0

Radioactive cascade
hope it helped!

You might be interested in

I know how to do electron configuration, but I think I’m doing the rest wrong. Answers and explanations would be much appreciate

Natasha_Volkova [10]

Your answers seem great so far, except for a tiny issue: With the ionic symbols, try to get into the habit of using "+", with metals, like sodium, and try to use the integer first. So, for example, a potassium ion would be K^+, while an oxide ion would be O^2-

Let's take aluminium as an example I'll work through:

Aluminium, with it's atomic number of 13, will have an electronic configuration of 1s2 2s2 2p6 1s2 2p1. So it would have 2, 8, 3 electrons in the first three energy levels, respectively.

Usually, if an elemental atom has a valence electron (highest energy level electron) count less than 4, it almost always will lose electrons. Since aluminium has 3, it will also lose the electrons.

It loses the 3 valence electrons, and so will end up with 10 electrons.

Since the atomic number also tells how many protons it has, we know that an aluminium atom has 13 protons, which doesn't change.

Since the size of the charges of a proton and an electron are the same, with protons being positive and the electrons being negative, an aluminium ion would have a charge of +3, and the Ionic symbol would be Al^3+

Hope I helped! xx

4 0

3 years ago

How many electrons in each orbit?

sveticcg [70]

In 1st orbit 2
2nd 8
3rd 10
f orbital has 16

4 0

3 years ago

Is mixing baking soda and vegetable oil a chemical change or physical change and WHY? PLEASE HELP

skelet666 [1.2K]

It’s a chemical change

3 0

2 years ago

Read 2 more answers

What phase of matter is least common on earth

Diano4ka-milaya [45]

Plasma is the least common matter on earth

4 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago