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stiks02 [169]
3 years ago
5

Compounds retain most of the properties of their constituent elements. True False

Chemistry
2 answers:
Pavlova-9 [17]3 years ago
5 0
Compounds retain most of the properties of their constituent elements. The correct answer is FALSE
vekshin13 years ago
3 0

Answer:

The correct answer is False.

Explanation:

Compounds do not retain the properties of their constituent elements. This is because the elements that form the new compounds have different bonds between them, and the intermolecular forces produce that change the physical and chemical properties of these new compounds. The state of aggregation at ambient temperature (melting and boiling points), electrical conductivity, hardness and solubility in polar and non-polar solvents depend on the strength of the chemical bond involved between the elements.

Have a nice day!

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Answer:

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Explanation:

4 0
2 years ago
Read 2 more answers
A solution contains some or all of the ions Cu2+,Al3+, K+,Ca2+, Ba2+,Pb2+, and NH4+. The following tests were performed, in orde
xeze [42]

Answer:

See below explanation

Explanation:

When having a mixture of metals in solution, you may perform an analytical study (using selective chemical conditions), that may help you to determine whether a metal (cation) is present or not

Using selective analytes (or conditions), leads to consecutive precipitations, until most of the cations are separated in precipitates

With this technique, you may identify metals in different groups, each group will have its analyte (or condition), which will help to have a different precipitate:

- Group I: Ag⁺, Pb⁺², Hg⁺²;  Analyte: HCL ; Precipitate: AgCl (white) , PbCl₂, HgCl₂

- Group II: As⁺³ , Bi⁺³, Cd⁺², Cu⁺² , Sb⁺³, Sn⁺² ; Analyte: H₂S (g) with HCL ; Precipitate: As₂S₃ , Bi₂S₃ , CdS (yellow) , CuS (black), Sb₂S₃, SnS

- Group III: Co⁺², Fe⁺², Fe⁺³, Mn⁺², Ni⁺², Zn⁺², Al⁺³, Cr⁺³; Analyte: NaOH or NH₃ with (NH₄)₂S (ac) ; Precipitate: CoS (black) , FeS, MnS , NiS (black), ZnS (white) , Al(OH)₃ (white), Cr(OH)₃  

- Group IV: Mg⁺², Ca⁺², Sr⁺², Ba⁺²; Analyte: Na₂CO₃ (ac) or (NH₄)₂HPO₄ (ac); Precipitate: respective carbonate or phosphate MgCO₃/MgHPO₄, CaCO₃/CaHPO₄ , SrCO₃/SrHPO₄, BaCO₃/BaHPO₄

- Group V: Li⁺, K⁺, Na⁺, Rb⁺, Cs⁺, NH₄⁺ ; will remain all in final solution

According to the original statement:

A solution contains one or more of the following: Cu⁺², Al⁺³, K⁺, Ca⁺², Ba⁺², Pb⁺², NH₄⁺

1) Addition on HCl 6M produces no change: we can say the sample does not contain Pb⁺² (group I)

2) Addition of H₂S with 0.2 M HCL produced a black solid: we could say sample contains Cu⁺²(group II)

3) Addition of (NH₄)₂HPO₄ in NH₃ produces no reaction: we could say we don´t have Ca⁺² and /or Ba⁺²  (group IV)

4) The final supernatant, when heated produced a purple flame: in the final solution, we have K⁺ (group V), which produces a purple flame (based on its characteristic emission spectrum when subjected to flame)

This analysis will be inconclusive for NH₄⁺ (according to above describe technique)

6 0
3 years ago
Why would you need to heat the hydrate up to 250 degrees if water boils at 100 degrees Celsius? Do you think that the boiling po
sergeinik [125]

Explanation:

The pure form of water has a boiling point of 100°C. Boiling point is a physical property of matter and it shows that at such temperature, a liquid substance will change state to vapor.

Pure water is made up of 2 atoms of hydrogen and 1 atom of oxygen. The only intermolecular forces between them is the hydrogen bonds that must be broken for the water to boiling off.

In hydrate, water is present in another form. The water is attached to another compound.

For a pure liquid, the they have reasonably constant boiling point and low boiling range.

Impurities such as the other part of the hydrate causes the elevation of the boiling point and the widening of the boiling range for impure substances.

We are no longer dealing with just hydrogen bonds, other molecular interactions are now involved and they need to be accounted for.

learn more:

Pure substances brainly.com/question/1832352

#learnwithBrainly

8 0
3 years ago
What types of intermolecular forces are found in CH<br> 4
Sphinxa [80]
London dispersion forces
8 0
3 years ago
6 Fe2+ (aq) + Cr2O72− (aq) + 14 H+ (aq) → 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (aq) If the titration of 23 mL of an iron(II) soluti
ratelena [41]

Answer:

1.047 M

Explanation:

The given reaction:

6Fe^{2+}_{(aq)}+Cr_2O_7^{2-}_{(aq)}+14H^+_{(aq)}\rightarrow 6Fe^{3+}_{(aq)}+2Cr^{3+}_{(aq)}+7H_2O_{(aq)}

For dichromate :

Molarity = 0.254 M

Volume = 15.8 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.8 ×10⁻³ L

Thus, moles of dichromate :

Moles=0.254 \times {15.8\times 10^{-3}}\ moles

Moles of dichromate = 0.0040132 moles

1 mole of dichromate react with 6 moles of iron(II) solution

Thus,

0.0040132 moles of dichromate react with 6 × 0.0040132 moles of iron(II) solution

Moles of iron(II) solution = 0.02408 moles

Volume = 23 mL = 0.023 L

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

<u>Molarity = 0.02408 / 0.023 = 1.047 M</u>

7 0
2 years ago
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