All categories

3 years ago

Identify two metalloids that have the same number of electrons shells krypton

Chemistry

1 answer:

disa [49]3 years ago

3 0

Answer;

Ge and As.

Explanation;

Elements Ge and As belong in the same period as krypton.

Each row of the periodic table is called the period.

Elements in each period have the same number of electron shells.

These elements in the same period do not have similar properties- the properties change greatly across every period. Krypton is in period four therefore all the elements in period 4 have four electrons shells.

You might be interested in

Which of these elements exhibits chemical behavior similar to that of oxygen? select one:

scoray [572]

Answer:
Sulfur (Option-C) <span>exhibits chemical behavior similar to that of oxygen.

Explanation:
Sulfur has same chemical properties as that of Oxygen because both of them belongs to same group in the periodic table. Also, the similarity of chemical behaviour among the group members is due to same number of electrons in their valence shells.
For examole, the electronic configuration of Oxygen is,

1s</span>², 2s², 2p⁴

There are six valence electrons in the valence shell (i.e. 2) of Oxygen.

Now for Sulfur,
1s², 2s², 2p⁶, 3s², 3p⁴

There are six valence electrons in the valence shell (i.e. 3) of Sulfur.

Therefore, both elements tends to gain 2 electrons in a reaction and form O⁻² and S⁻² respectively.

3 0

3 years ago

For question 9 I need help push or pull

adelina 88 [10]

Answer:

a push

please thank and brainlest!

Explanation:

5 0

2 years ago

You measure the absorbance of your extracted curcumin solution and realize that the solution is too concentrated. What made you

zepelin [54]

If the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high.

Spectrometry measures the interaction of light with molecules. The absorbance refers to how much light that interacts with molecules of the substance. The more the concentration of the substance the higher the absorbance of the solution.

Hence, if the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high. An unusually high absorbance tells us that the solution is too concentrated.

Learn more: brainly.com/question/13440572

8 0

2 years ago

If a wave has a frequency of 2.85x10^10 its wavelength is ?

Ede4ka [16]

Answer:

tepo jds fnsdmv dfjv dfj pvjdf ifjv dfjv sdijghdfuiExplanation:

3 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