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12345 [234]
3 years ago
15

A beam of light passes though a liquid in a test tube without scattering. Which type of mixture is most likely in the test tube?

Chemistry
2 answers:
8_murik_8 [283]3 years ago
8 0

<u>Answer:</u> The correct answer is Option C.

<u>Explanation:</u>

Colloid is defined as the solution where particle size is between 2 to 1000 nm.  In these solutions, a physical distinction between the particle is seen. This solution can scatter light when a beam of light is passed through it.

Emulsion is defined as fine dispersion of droplets of one liquid in another liquid in which it is not soluble. There is a physical distinction between the particles and thus they can scatter the light falling on them.

True solution is defined as the solution where the size of the particle is between 0.1 nm to 1 nm. There is no physical distinction between the particles and thus they cannot scatter the light falling on them.

Suspension is defined as the solution in which the size of the particle is greater than 1000 nm. There is physical distinction between the particles and thus they can scatter the light falling on them.

Hence, the correct answer is Option C.

butalik [34]3 years ago
6 0
The answer would be letter C - solution.

A mixture should be homogeneous for a light not to be scattered. This is because particles are distributed evenly throughout the mixture which allows light to pass directly. In your choices, the solution allows a  beam of light to pass through a liquid in a test tube without scattering.
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What intermolecular force attracts two non polar molecules to each other
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The intermolecular force that attracts two nonpolar molecules is London dispersion forces, which are also called induced dipole-induced

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2 years ago
Is air homogeneous or heterogeneous?
ivolga24 [154]

Answer:  yes, Air is a homogeneous mixture of the gaseous substances nitrogen, oxygen, and smaller amounts of other substances. Salt, sugar, and substances dissolve in water to form homogeneous mixtures. A homogeneous mixture in which there is both a solute and solvent present is also a solution

Explanation:

8 0
2 years ago
Read 2 more answers
Determine the molecular
jeyben [28]

Answer:

MgO- magnesium oxide

Cu(NO3)2- copper(11)nitrate

Li2CO3- lithium carbonate

7 0
3 years ago
Many chemical names are similar at first glance. Give the formulas of the species in each set: (a) ammonium ion and ammonia; (b)
ohaa [14]

Answer:

a) Ammonium ion: NH₄⁺, ammonia: NH₃.

b)Magnesium sulfide: MgS, magnesium sulfite: MgSO₃, magnesium sulfate: MgSO₄.

c) Chloric acid: HClO₃, chlorous acid HClO₂.

d) Cupric bromide: CuBr₂, cuprous bromide: CuBr.

Explanation:

a) Ammonia is the neutral substance, which as molecular formula NH₃ when it gains a proton, it becomes the ammonium ion NH₄⁺.

b) In the name of salts, first, it comes the cation name, in this case, all the cations are the magnesium (Mg²⁺). The ions of sulfur can be oxidized or not, the termination "ide" indicates the non-oxidized ion (S⁻²), "ite" the less oxidized (SO₃⁻²), and "ate" the more oxidized (SO₄⁻²). To do the molecular formula, the charges are replaced without the signals and put down. If they are equal, the number is not put.

Magnesium sulfide: MgS

Magnesium sulfite: MgSO₃

Magnesium sulfate: MgSO₄

c) The acids can be oxidized or not, when it has no oxygens, the name is given "hydro" + the anion name, as hydrochloric acid (HCl). When there's oxygen in the molecule, the more oxidized is name as "'per" + anion name + "ic", the second more oxidized as anion name + "ic", the second less oxidized as anion name + "us", and the less oxidized as "hypo" + anion name + "us". In this case:

chloric acid: HClO₃

chlorous acid HClO₂

d) The salts of copper are named depends on the oxidation number of the cupper. The one with a higher oxidation number( +2) is called cupric, and the other (+1), cuprous. Thus:

cupric bromide: CuBr₂ (oxidation number = +2)

cuprous bromide: CuBr (oxidation number = +1)

5 0
2 years ago
Read 2 more answers
What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?
asambeis [7]
Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Putting Values,
 
                            W  =  (10 A × 30.6 s × 31.77 g) ÷ 96500

                            W  =  0.100 g

Result:
           0.100 g of Cu
²⁺ is deposited.
3 0
3 years ago
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