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alekssr [168]
3 years ago
9

Why is a solution not a pure substance

Chemistry
1 answer:
Firlakuza [10]3 years ago
8 0
It is not a pure substance, because a solution are mixed chemicals in a way that the molecules are not bonded with one another. Thus, separating them from compounds and elements, which are pure substances.
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Use the pictures below to answer the following question what type of bond is this
mezya [45]
I literally hate chem but I think it’s ionic, I’m not not completely sure but it kinda sounds about right . Not the best anwser haha hope it kinda helps lol
4 0
3 years ago
An alkene X undergoes ozonolysis and gives two compounds Y and Z of molecular formula CaHO. Y and Z are functional isomers of ea
lys-0071 [83]

i. The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.

ii. The structural formula of Y is CH_3-CO-CH_3 and Z is CH_3-CH_2-CHO.

iii. Alkenes, upon catalytic hydrogenation, form alkanes. This will occur in the presence of Nickel as the catalyst.

iv. The process of ozonolysis is useful in the field of pharmaceutics.

v. The test of unsaturation can be performed by passing a compound through Bromine water.

<h3>What is ozonolysis?</h3>

Ozonolysis is a reaction used in organic chemistry to determine the position of a carbon-carbon double bond in unsaturated compounds.

i. The given alkene X, that is subject to ozonolysis would be 2-methyl-2-pentene. Upon exposure to ozone, an ozonide is initially formed, after which it is broken down into 2 products - acetone and propanal, both with the molecular formula C₃H₆O.

The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.

ii. The formulas of Y is CH_3-CO-CH_3 and Z is CH_3-CH_2-CHO. They are functional isomers as they have the same molecular formula but different functional groups - ketone and aldehyde.

iii. When alkenes undergo catalytic hydrogenation, they form alkanes, X will form 2 methyl petane on reaction with hydrogen gas in presence of Ni.

iv. The ozonolysis is used for the industrial-scale synthesis of pharmaceuticals.

v. The unsaturation of compound X can be proved by the bromine water test. As on reaction with it, the brown colour of bromine water becomes colourless due to the formation of dibromo alkane.

Learn more about the ozonolysis here:

brainly.com/question/14356308

#SPJ1

5 0
2 years ago
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera
Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0
3 years ago
The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0
2 years ago
Which of the following terms describes an accumulation of rocky, sandy, or clayey material deposited at the end of a glacier?
OLEGan [10]

Answer:

Option (D)

Explanation:

<u>Eskers are the long ridges that are comprised of rocks, sands and clay particles and are deposited towards the end of the glaciers</u>. These are fluvioglacial depositional features. These particles are exposed after the glaciers recede. These ridges are formed parallel to the earlier flow direction of ice. The size of eskers is generally smaller as it carries smaller particles such as rocks, sands, and gravels, in comparison to the different type of moraines. It is because the flow velocity decreases as the glaciers melt. So, these eskers are formed at the end of the glaciers.

Thus, the correct answer is option (D).

3 0
3 years ago
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