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3 years ago

Why is the separation of mixtures into pure or relatively pure substances so important when performing a chemical analysis?

Chemistry

1 answer:

N76 [4]3 years ago

3 0

Answer:

It is important to separate mixture into pure or relatively pure substances when performing a chemical analysis SO AS TO KNOW THE PROPERTIES COMING FROM EACH PART MIXTURE WHICH MAY INTERFERE WITH THE SEPARATION.

Explanation:

In chemistry, Mixture is the combination of two or more substances which are not combine chemically.

Mixture contain different substances with different physical and chemical properties.

It is important to purify the substances in a mixture so as to identify what properties are coming from each mixture and also some part of the mixture can interfere with the properties of other mixture present for skewing analysis.

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PLZ HELP IM CONFUSED

Marina CMI [18]

Answer:

9.1 mol

Explanation:

The balanced chemical equation of the reaction is:

CO (g) + 2H2 (g) → CH3OH (l)

According to the above balanced equation, 2 moles of hydrogen gas (H2) are needed to produce 1 mole of methanol (CH3OH).

To convert 36.7 g of hydrogen gas to moles, we use the formula;

mole = mass/molar mass

Molar mass of H2 = 2.02g/mol

mole = 36.7/2.02

mole = 18.17mol

This means that if;

2 moles of H2 reacts to produce 1 mole of CH3OH

18.17mol of H2 will react to produce;

18.17 × 1 / 2

= 18.17/2

= 9.085

Approximately to 1 d.p = 9.1 mol of methanol (CH3OH).

6 0

2 years ago

. In which reaction is nitric acid acting as an oxidising agent?

Talja [164]

Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

Hello,

In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.

Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.

Regards.

8 0

2 years ago

Read 2 more answers

Consider three gases: Ar, SF6, and Cl2. If 50.0 grams of these gases are placed in each of three identical containers, which con

adoni [48]

The ideal gas law:
$pV=nRT \Rightarrow p=\frac{nRT}{V}$
p - pressure, n - number of moles, R - the gas constant, T - temperature, V - volume

The volume and temperature of all three containers are the same, so the pressure depends on the number of moles. The greater the number of moles, the higher the pressure.
The mass of gases is 50 g.

$Ar \\
M \approx 39.948 \ \frac{g}{mol} \\
n=\frac{50 \ g}{39.948 \ \frac{g}{mol}} \approx 1.25 \ mol \\ \\
SF_6 \\
M \approx 146.06 \ \frac{g}{mol} \\
n=\frac{50 \ g}{146.06 \ \frac{g}{mol}} \approx 0.34 \ mol \\ \\
Cl_2 \\
M=70.9 \ \frac{g}{mol} \\
n=\frac{50 \ g}{70.9 \ \frac{g}{mol}} \approx 0.71 \ mol$

The greatest number of moles is in the container with Ar, so there is the highest pressure.

4 0

3 years ago

What is the first step of seafloor spreading?

drek231 [11]

Answer:

1. A long crack in the oceanic crust forms at a mid ocean ridge

Explanation:

7 0

2 years ago

Consider the following three-step reaction pathway.

Alenkinab [10]

Answer:

$NO \longrightarrow N_2O_2 \longrightarrow N_2O \longrightarrow N2$

Explanation:

The intermediates are the products of all the steps of the reaction pathway, with the exception of the last one. So the intermediates will be:

N2O2 from the first step
N2O from the second step

The list from reactant to final product:

$NO \longrightarrow N_2O_2 \longrightarrow N_2O \longrightarrow N2$

<em>Note: the water is considered a by-product, given that is not the product of interest in this steps.</em>

7 0

3 years ago