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soldier1979 [14.2K]
3 years ago
9

Which element is most chemically similar to chlorine? Ar F Fr S

Chemistry
1 answer:
ElenaW [278]3 years ago
3 0
All halogens
 for this F is correct
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A colloid can be detected by using the _________ effect. mohs tyndall faraday brownian
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How to make Sodium thiosulfate
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Explanation:

Method of prepration of sodium thiosulphate - definition

In the laboratory, this salt can be prepared by heating an aqueous solution of sodium sulphite with sulphur or by boiling aqueous NaOH and sulfur according to this equation:

6NaOH + 4 S _{2} Na _{2} S ->Na _{2} S_{2}O_{3}+ 3H _{2}O.

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3 years ago
Select the following statements that are true. Select all that apply.
geniusboy [140]

Answer:

The statements which are true among these are: (a),(b) and (c) because,

(a) The simplest organic compounds which contains only carbon and hydrogen atoms are called hydrocarbons.

(b) The IUPAC naming of organic compounds have some rules for the naming of compounds, which consists of

  • Finding the longest chain present in the compound called parent chain.
  • A prefix for any substituent attach to the parent chain.

And lastly a suffix for the type of bond that molecule have.

(c) Isomers are the compound which same same molecular formula but different arrangement of molecules, due to this different arrangement they have different physical and chemical properties.

4 0
4 years ago
15g of FeCI3 is dissolved in 450 mL of solution. What is the concentration of [CI-]?
storchak [24]

The concentration of [CI-] : 0.617 M

<h3>Further explanation</h3>

FeCl₃ dissolved in 450 mL of solution(will dissociate )

Reaction

FeCl₃⇒Fe³⁺+3Cl⁻

  • mol FeCl₃(MW=162,2 g/mol)

\tt \dfrac{15}{162.2}=0.0925

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\tt \dfrac{3}{1}\times 0.0925=0.2775

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7 0
3 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
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