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konstantin123 [22]
2 years ago
6

Investigation of ions and Separation of Mixtures 1- Propose systematic schemes for the identification of the following salts wit

h respect to their anions and cations: NaCl, KNO3, and CuSO4?​
Chemistry
1 answer:
zalisa [80]2 years ago
5 0

The cations and anions can be identified using cataloged reactions schemes. For instance, the copper II ion can be identified by reaction with sodium hydroxide.

The sodium cation is easily identified by flame test. sodium imparts a yellow color to a flame. The chloride ion is identified by the use of a mixture of HNO3/AgNO3 solution. The color of the precipitate shows which halide ion is present. A white precipitate indicates the presence of the chloride ion.

The potassium cation is also identified by flame test. The ion imparts a lilac color to flame. Addition of acidified FeSO4 solution is used to confirm the presence of the nitrate ion. Formation of a brown ring is a positive test for the nitrate ion.

For CuSO4, the presence of copper II ion can be confirmed using dilute NaOH. If a light blue precipitate is formed which dissolves in excess NaOH then the copper II ion is confirmed. The presence of the sulfate ion is confirmed using a solution of barium nitrate and dilute nitric acid. Formation of a white precipitate is a positive test for the sulfate ion.

Learn more: brainly.com/question/5624100

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Answer:

We know that,

Density= mass upon volume.

mass= 21.93g

Volume= 49.3 - 46.3

= 3 cm³

Density = 21.93÷ 3

= 7.31 g/cm³

Hope it helps :))

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What volume is occupied by 16.4 g of mercury? The density of mercury is 13.6 g/mL
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3 years ago
As you read from left to right across the periodic table atomic numbers and what by one of each element
Illusion [34]

Answer:

As we read from left to right across the periodic table atomic numbers are increased by one each of element.

Explanation:

As we move from left to right across the periodic table the atomic number is increased by one and the number of valance electron in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

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3 years ago
What are the properties of salt
GrogVix [38]
The compound is sodium chloride 
6 0
3 years ago
Read 2 more answers
The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
enot [183]

Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

(b) Moles of CO₂

\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

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