All categories

1 year ago

Is Chemistry the study of the submicroscopic, the microscopic, the macroscopic, or all three?

Chemistry

1 answer:

alexandr402 [8]1 year ago

4 0

The question requires us to comment on what are the subjects studied by Chemistry.

Chemistry is the study of all matter and the transformations that this matter undergo. As matter exists in macroscope (all of we can see, touch and measure directly), microscope (the small particles that compose all matter) and submicroscope (the smallest particles that make up the microscope particles) forms, we can say that Chemistry englobes the study of macroscope, microscope and submicroscope matter.

You might be interested in

At 1000 K, a sample of pure NO2 gas decomposes. 2 NO2(g) equilibrium reaction arrow 2 NO(g) + O2(g) The equilibrium constant KP

kirill115 [55]

<u>Answer:</u> The pressure of NO and $NO_2$ in the mixture is 0.58 atm and 0.024 atm respectively.

<u>Explanation:</u>

We are given:

Equilibrium partial pressure of $O_2$ = 0.29 atm

For the given chemical equation:

$2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)$

Initial: a

At eqllm: a-2x 2x x

Calculating for the value of 'x'

$\Rightarrow x=0.29$

Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm

Equilibrium partial pressure of $NO_2$ = a - 2x = a - 2(0.29) = a - 0.58

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{O_2}\times (p_{NO})^2}{(p_{NO_2})^2}$

We are given:

$K_p=158$

Putting values in above expression, we get:

$158=\frac{0.29\times (0.58)^2}{(a-0.58)^2}\\\\a=0.555,0.604$

Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.

So, $a=0.604$

Equilibrium partial pressure of $NO_2$ = (a - 0.58) = (0.604 - 0.58) = 0.024 atm

Hence, the pressure of NO and $NO_2$ in the mixture is 0.58 atm and 0.024 atm respectively.

4 0

3 years ago

How many grams of carbon dioxide are in 88.5 L at STP?

34kurt

Answer:

173.8g

Explanation:

STP means standard temperature and pressure

The temperature there is 273k while the pressure is 1 atm

now we are to use the ideal gas equation to get the number of moles first

Mathematically;

PV = nRT

here P = 1 atm

V = 88.5 L

n = ?

R = molar gas constant = 0.082 L atm mol^-1 K^-1

Now rewriting the equation we can have

n = PV/RT

plugging the values we have

n = (1 * 88.5)/(0.082 * 273)

n = 88.5/22.386

n = 3.95 moles

Now we proceed to get the mass

Mathematically;

mass = no of moles * molar mass

molar mass of carbon iv oxide is 44g/mol

mass = 3.95 * 44 = 173.8 g

6 0

3 years ago

A mixture in which the suspended particles are too small to be seen with the naked eye is a _____.

Degger [83]

B) colloid is the correct answer

6 0

3 years ago

Read 2 more answers

3.15.8 cm3 of H2SO4 completely neutralised 20 cm3 of NaOH. The NaOH was 1.5 mol/dm3.

Furkat [3]

Answer:

Molarity = 0.95 mol/dm³

Explanation:

Given data:

Volume of H₂SO₄ = 15.8 cm³

Volume of NaOH = 20 cm³

Concentration of NaOH = 1.5 mol/dm³

Concentration of H₂SO₄ = ?

Solution:

Chemical equation:

NaOH + H₂SO₄ → Na₂SO₄ + H₂O

First of all we will calculate the number of moles of NaOH and for that we will convert the units first,

Volume = 20 cm³/1000 = 0.02 L

Concentration of NaOH = 1.5 mol/dm³

1 mol/dm³ = 1 mol/L

Concentration of NaOH = 1.5 mol/L

Number of moles of NaOH:

Molarity = number of moles / volume in L

1.5 M = number of moles / 0.02 L

Number of moles = 1.5 M ×0.02 L

Number of moles = 0.03 mol

Now we will compare the moles of NaOH and H₂SO₄

NaOH : H₂SO₄

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Concentration of H₂SO₄:

Volume of H₂SO₄:

15.8 cm³/1000 = 0.0158 L

Molarity = number of moles / volume in L

Molarity = 0.015 mol / 0.0158 L

Molarity = 0.95 mol/L

1 mol/L = 1 mol/dm³

Molarity = 0.95 mol/dm³

6 0

3 years ago

30 Points!

olga55 [171]

I would imagine it’s cleaning the crime scene after evidence has been collected. they do so much more but out of these, the first one is the best answer.

4 0

3 years ago