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deff fn [24]
2 years ago
8

How do you know something is there if you can't see it?

Chemistry
2 answers:
gavmur [86]2 years ago
5 0

Answer:

we know that someone or something is near us even when we cant see it you see our self conscious minds can pick up on the slightest detail even when we dont pick up on it also you can  thank the six senses as well for why we can tell something or someone is there

Explanation: i watch criminal minds and pay attention to the random facts spencer reid spits out

aliina [53]2 years ago
3 0

Answer:

Explanation:

Use your other senses such as touch or smell, does it react in different temperatures, is it then visible under a microscope, does it react with different elements or gases?   Think outside the box.

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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
What are the spectator ions in this reaction
Lady_Fox [76]

sodium ions and chloride ions

6 0
3 years ago
Which of the following atoms would gain two electrons to fill its valence energy level?
Harlamova29_29 [7]

The atom that would gain two electrons to fill its valence energy level is S(sulfur)

This is because s (sulfur) is in atomic number 16 with 2.8.6 of [Ne] 3s^2 2p^4 electronic configuration. This implies that sulfur has 6 valence electron and therefore it require two electron to fill its valence energy level and obtain 18 rule electrons.

5 0
3 years ago
Read 2 more answers
1. To calculate the molarity of a solution, you need to know the moles of solute and the
evablogger [386]

Answer:

O volume of the solution

Explanation:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / volume of solution  in L

For example:

if we dissolve the 1 mole of NaCl to make the solution of volume 2 L. The molarity of solution is,

M = 1 mol / 2 L

M = 0.5 M

3 0
3 years ago
For the balanced chemical reaction
musickatia [10]

Answer:

150

Explanation:

  • C₄H₂OH + 6O2 → 4CO2 + 5H₂O

We can <u>find the equivalent number of O₂ molecules for 100 molecules of CO₂</u> using a <em>conversion factor containing the stoichiometric coefficients of the balanced reaction</em>, as follows:

  • 100 molecules CO₂ * \frac{6moleculesO_2}{4moleculesCO_2} = 150 molecules O₂

150 molecules of O₂ would produce 100 molecules of CO₂.

5 0
2 years ago
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