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Tom [10]
3 years ago
8

A compound is broken down by chemical means during

Chemistry
2 answers:
alexandr1967 [171]3 years ago
6 0

Answer : Option 3) Electrolysis.


Explanation : In the process of electrolysis, decomposition of the chemical is done by passing the electric current from the liquid or the solution containing ions in it. This is the process where compound is broken into ions namely, cations and anions.


Rest all options does not involves chemical decomposition of the compound.


Chromatography is a technique where rates of solute and solvent are compared in a medium where the solute components move at different rates and the separation is obtained.


Distillation simply purifies the liquid or solution by heating or cooling process.


Filtration just involves removal of impurities by using a more porous material device to get purified compound.

ELEN [110]3 years ago
3 0

Answer:

The correct answer is option (3).

Explanation:

Electrolysis is defined as chemical process in which decomposition is carried out by passing an electricity in a solution containing ions. For example: electrical decomposition of water gives oxygen  gas and hydrogen gas.

2H_2O(l)\overset{Electricity}\rightarrow 2H_2(g)+O_2(g)

where as chromatography, distillation and filtration are the separation techniques used to separate the compounds from the mixtures.

Hence, the correct answer is option (3).

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The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
3 years ago
A six-carbon organic compound containing oxygen is suspected of being either a secondary alcohol or a ketone. which chemical or
White raven [17]
Following are the possible isomers of secondary alcohol and ketones for six carbon molecules. In order to distinguish between sec. alcohol and ketone we can simply treat the unknown compound with acidified Potassium Dichromate (VI) in the presence of acid. If with treatment with unknown compound the colour of K2Cr2O7 (potassium dichromate VI) changes from orange to green then it is confirmed that the unknown compound is sec. alcohol, or if no change in colour is detected then ketone is confirmed. This is because ketone can not be further oxidized while, sec. alcohol can be oxidized to ketones as shown below,

6 0
3 years ago
Who believed that a grain of sand could be divided indefinitely?
Ilya [14]

Answer:

Democritus

Explanation:

4 0
3 years ago
Find the height of a triangle that has an area of 30 square units and a base measuring 12 units.
Gala2k [10]
The answer is b. 5 units
5 0
3 years ago
Assume that the density and heat of combustion of E85 can be obtained by using 85 % of the values for ethanol and 15 % of the va
Alika [10]

Answer:

87.4 J

Explanation:

The density of the gasoline is 0.70 g/mL, and the density of the ethanol is 0.79 g/mL. The heat combustions (the heat released in a combustion reaction) are 5,400 kJ/mol for gasoline, and 1,370 kJ/mol for ethanol.

For 3.5 L of E85, the volumes of gasoline and ethanol are:

Vgasoline = 0.15 * 3.5 = 0.525 L = 5.25x10⁻⁴ mL

Vethanol= 0.85 * 3.5 = 2.975 L = 2.975x10⁻³ mL

The mass of gasoline and ethanol presented in that sample of E85 is the volume multiplied by the density:

mgasoline = 5.25x10⁻⁴ * 0.70 = 3.675x10⁻⁴ g

methanol = 2.975x10⁻³ * 0.79 = 2.35025x10⁻³ g

The number of moles for each substance is it mass divided by its molar mass. The molar masses are 114 g/mol for gasoline, and 46 g/mol for ethanol:

ngasoline = 3.675x10⁻⁴/114 = 3.224x10⁻⁶ mol

nethanol = 2.35025x10⁻³ /46 = 5.109x10⁻⁵ mol

The energy released is the heat combustion multiplied by the number of moles, so:

Egasoline = 5,400 * 3.224x10⁻⁶ = 0.0174 kJ = 17.4 J

Eethanol = 1,370 * 5.109x10⁻⁵ = 0.07 kJ = 70 J

So, the energy released by the E85 is the sum of the energy released by ethanol and gasoline:

The energy released by E85 = 87.4 J

8 0
3 years ago
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