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3 years ago

Which list of elements from Group 2 on the Periodic Table is arranged in order of increasing atomic radius?

1 answer:

4 0

The correct answer is option 1. Be, Mg, and Ca is the correct order arranged in increasing atomic radius. This is predicted based on the periodic table. The atomic sizes increases as one moves downwards in the periodic table.

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A buffer is prepared by mixing 50.3 mL of 0.183 M NaOH with 128.8 mL of 0.231 M acetic acid. What is the pH of this buffer

kvv77 [185]

Answer:

A buffer system can be made by mixing a soluble compound that contains the conjugate ... 10.0 grams of sodium acetate in 200.0 mL of 1.00 M acetic acid.

Explanation:

7 0

2 years ago

Please help What is the possible ions of vanadium?

svetlana [45]

V+5, V+4, V+3, and V+2

7 0

3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

explain using diagrams how potassium forms the compound potassium flouride when it reacts with flourine

Katarina [22]

Answer:

The answer to your question is given below.

Explanation:

Potassium (K) has 19 electrons with electronic configuration of 2, 8, 8, 1.

Fluorine (F) has 9 electrons with electronic configuration of 2, 7.

Fluorine needs 1 electron to complete it's octet configuration.

Hence, potassium (K), will lose 1 electron to fluorine (F) to form potassium ion (K+) with electronic configuration of 2, 8, 8. The fluorine atom (F) will receive the 1 electron from potassium to form the fluoride ion (F-) with electronic configuration of 2, 8.

**** Please see attached photo for further details.

7 0

3 years ago

What is the transfer of electrons in Al + Cl = AlCl3

otez555 [7]

Answer:

3 e⁻ transfer has occurred.

Explanation

This is a redox reaction.

Oxidation (loss of electrons or increase in the oxidation state of entity)

Reduction (gain of electrons or decrease in the oxidation state of the entity)
An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
[Ne]= (1s²) (2s² 2p⁶)

A combination of both the reactions( Half-reactions) leads to a redox reaction.

Let us look at initial configurations of Al and Cl

[Al]= 1s² 2s² 2p⁶ 3s² 3p¹

[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵

Hence, Al can lose 3 electrons to achieve octet config.

and, Cl can gain 1e to achieve nearest noble gas config. [Ar]

This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.

Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃

Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)

Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.

3 0

3 years ago