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natima [27]
3 years ago
13

How should a segregated heterogeneous material be sampled in order to construct a representative sample

Chemistry
1 answer:
NikAS [45]3 years ago
7 0

First, separate them into homogenous units and then randomly select representative samples from each unit.

<h3>Heterogeneous population and sampling</h3>

In order to sample a heterogenous population or material and we want the sample to be representative of the population or material:

  • The material should first be separated into homogenous units
  • Each unit can then be randomly sampled or any other sampling method that is unbiased

More on heterogeneous populations can be found here: brainly.com/question/1068437?referrer=searchResults

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In the following reaction, identify the oxidized species, reduced species, oxidizing agent, and reducing agent. Be sure to answe
gtnhenbr [62]

Answer :

Cl_2 is reduced species.

KI is oxidized species.

Cl_2 is oxidizing agent.

KI is reducing agent.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The balanced redox reaction is :

Cl_2(aq)+2KI(aq)\rightarrow 2KCl(aq)+I_2(aq)

The half oxidation-reduction reactions are:

Oxidation reaction : 2I^-\rightarrow I_2+2e^-

Reduction reaction : Cl_2^++2e^-\rightarrow 2Cl^-

From this we conclude that the 'KI' is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and 'Cl_2' is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.

Thus, Cl_2 is reduced species.

KI is oxidized species.

Cl_2 is oxidizing agent.

KI is reducing agent.

7 0
4 years ago
What mass of sodium chloride contains 4.59 x 10 24 formula units?
shtirl [24]
The term formula units means molecules.

Then, what you are looking for is the mass in 4.59*10^24  molecules.

The procedure involves to convert the 4.59 * 10^24 molecules into moles and use the molar mass of the sodium chloride.

1) Number of moles = 4.59 * 10^24 molecules / (6.02 * 10^23 molecules/mol) = 7.62 mol

2) Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

3) mass of NaCl = molar mass *  number of moles = 58.44 g/mol * 7.62 mol = 445.31 g of NaCl

Answer: 445.31 g of NaCl.

 
7 0
3 years ago
Malik analyzed three samples to test which element was a metalloid. The table shows his results. A 7-column table with 3 rows. T
Maru [420]

Answer:C is most likely a metalloid

Explanation:A is a metal while B is a non metal. This is because when we look at the properties of metals, they are very lustrous and have a high conductivity. On the other hand, non metals are the opposite but metalloids have properties the are inbetween ie, they react with either an acid or base, and have a medium electrical conductivity unlike metals(high) anc non metals(low).

6 0
3 years ago
Read 2 more answers
CsH«N2(l) + CuCl(s)
bulgar [2K]

0.498 moles of copper(II) phthalocyanine would be produced by

the complete cyclotetramerization of 255 grams of phthalonitrile in the

presence of excess copper(ll) chloride.

<em>Copper(ll) phthalocyanine (Cu(C₃₂H₁₆N₈)) is produced by the cyclotetramerization of phthalonitrile (C₈H₄N₂) according to the following reaction: 4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g) How many moles of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 255 grams of phthalonitrile in the presence of excess copper(II) chloride?</em>

Let's consider the following balanced equation.

4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)

The molar mass of C₈H₄N₂ is 128.13 g/mol. The moles corresponding to 255 g of C₈H₄N₂ are:

255 g \times \frac{1mol}{128.13 g} = 1.99 mol

The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 1.99 moles of C₈H₄N₂ are:

1.99 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.498 mol Cu(C_{32}H_{16}N_8)

0.498 moles of copper(II) phthalocyanine would be produced by

the complete cyclotetramerization of 255 grams of phthalonitrile in the

presence of excess copper(ll) chloride.

You can learn more about stoichiometry here: brainly.com/question/22288091

8 0
2 years ago
What was one of the first series of event in the study of chemistry
baherus [9]
Discovered and isolated radium and polonium in 1898
7 0
3 years ago
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