Ask question

Ask question

All categories

3 years ago

11

A substance X is formed from an exhaustible natural resource . It is a black and thick liquid with an unpleasant smell .It is a

mixture of 200 substances.Some of the products such as dyes,explosives,paints,perfumes etc. Are obtained using this resource.The substance X is

1 answer:

mario62 [17]3 years ago

3 0

Answer:

crude oil

Explanation:

<em>The substance X would be crude oil.</em>

<u>Crude oil is a non-renewable natural resource that is black and thick in appearance. It is a mixture of several products and these are often separated by a process known as fractional distillation. </u>

Some of the products from the fractional distillation of crude oil include the premium motor spirit (petrol), automotive gas oil (diesel), kerosene, etc., and derivatives for the production of dyes, paints, perfumes, etc.

You might be interested in

]\ \textless \ br /\ \textgreater \ \huge \boxed{ \red{φµεรƭเσɳ -}} \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \te

azamat

Answer:

Nitrogen fixation is a chemical process by which molecular nitrogen in the air is converted into ammonia or related nitrogenous compounds in soil or aquatic systems. Atmospheric nitrogen is molecular dinitrogen, a relatively nonreactive molecule that is metabolically useless to all but a few microorganisms.

Explanation:

6 0

3 years ago

Read 2 more answers

A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.

Lilit [14]

Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

$\frac{V1}{T1} = \frac{V2}{T2}$

Solve for V2

V2 = $\frac{V1T2}{T1}$

3.- Substitution

V2 = $\frac{(42)(233)}{293}$

4.- Simplification

V2 = $\frac{9786}{293}$

5.- Result

V2 = 33.4ml

3 0

3 years ago

An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about 4.4x 10 m/s Calculate the

otez555 [7]

The question is incomplete, here is the complete question:

An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about $4.4\times 10^6m/s$ . Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of helium. Write your answer as a percentage of the average speed, and round it to 2 significant digits

<u>Answer:</u> The percentage of average speed is 41 %

<u>Explanation:</u>

We are given:

Radius of helium atom = 31 pm = $31\times 10^{-12}m$ (Conversion factor: $1m=10^{12}pm$ )

So, diameter of helium atom = $(2\times r)=(2\times 31\times 10^{-12})=64\times 10^{-12}m$

The diameter of the atom will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $64\times 10^{-12}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of electron = $9.1095\times 10^{-31}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$64\times 10^{-12}m\times 9.1095\times 10^{-31}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 64\times 10^{-12}m\times 9.1095\times 10^{-31}kg}=1.81\times 10^6m/s$

To calculate the percentage of average speed, we use the equation:

$\text{Percentage of the average speed}=\frac{\text{Uncertainty in velocity}}{\text{Average orbital speed}}\times 100$

We are given:

Average orbital speed = $4.4\times 10^6m/s$

Putting values in above equation, we get:

$\text{Percentage of the average speed}=\frac{1.81\times 10^6m/s}{4.4\times 10^6m/s}\times 100\\\\\text{Percentage of the average speed}=41.\%$

Hence, the percentage of average speed is 41 %

3 0

4 years ago

Which two electron configurations represent elements that would have similar chemical properties? (1) 1s22s22p4 (2) 1s22s22p5 (3

Anton [14]

Answer: Two electron configurations of elements that would have similar chemical properties are (2) and (4).

Explanation:

(1) $1s^22s^22p^4$

(2) $1s^22s^22p^5$

(3) $[Ar]4s^23d^5$

(4) $[Ar]4s^23d^{10}^4p^5$

Valence electrons : These are the electrons present in last principal quantum number of an atom of the element.

Two electron configurations represent elements that would have similar chemical properties are (2) and (4). This is because number of valence electrons present in both of them are same that is seven valence electrons . Also valence electrons of both the elements are present in p-orbital which means that they both belongs to same group in a periodic table. The members of same group in a periodic table have similar chemical properties.

3 0

3 years ago

Calculate the formula mass for the compound nitrous acid.

kari74 [83]

Answer:

47.01g/mol

Explanation:

The formula of Nitrous acid is HNO₂:

The formula mass of a compound gives the sum of the atomic masses of the elements that makes up a compound.

Atomic mass of H = 1.0079g/mol

N = 14.0067g/mol

O = 15.9994g/mol

Formula mass of HNO₂ = 1.0079 + 14.0067 + 2(15.9994)

= 1.0079 + 14.0067 +31.9988

= 47.01g/mol

8 0

4 years ago