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3 years ago

Classify these substances? More than one answer may apply in each case.

Chemistry

1 answer:

dalvyx [7]3 years ago

3 0

Answer:

N2 element, pure substance

O2 element, pure substance

N2O Compound, pure substance

Air (mostly N2 and O2 ) homogeneous mixture

Explanation:

N2, Nitrogen is known as the chemical element that is characterized by having atomic number 7 and that is symbolized by the letter N, in its molecular version, it is recognized as N2.

O2, Oxygen is the chemical element of atomic number 8, this molecular form is composed of two atoms of this element.

A chemical element is a type of matter, consisting of atoms of the same class.

N2O, Nitrous oxide is formed by the union of two molecules of nitrogen and one of oxygen, which is considered a chemical compound since it is a substance formed by the chemical combination of two different elements of the periodic table.

A pure substance is one that cannot change state or divide into other substances, except for a chemical reaction.

Air (mostly N2 and O2 ), it is a homogeneous mixture of gases that constitutes the earth's atmosphere. A homogeneous mixture is a type of mixture in which its components are not distinguished and in which the composition is uniform and each part of the solution has the same properties.

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The bromination of acetone is acid-catalyzed.CH3COCH3 + Br2 CH3COCH2Br + H+ + Br -The rate of disappearance of bromine was measu

Ann [662]

Answer:

a) The rate law is:

rate = k[Acetone][Br₂]⁰[H⁺] = k[Acetone][H⁺]

b) The value of k is:

k = 3.86 × 10⁻³ M⁻¹ · s⁻¹

Explanation:

Acetone (M) Br2 (M) H+ (M) Rate (M/s)

0.30 0.050 0.050 5.7 x 10-5

0.30 0.10 0.050 5.7 x 10-5

0.30 0.050 0.10 1.2 x 10-4

0.40 0.050 0.20 3.1 x 10-4

0.40 0.050 0.050 7.6 x 10-5

A generic rate law for this reaction could be written as follows:

rate = k[Acetone]ᵃ[Br₂]ᵇ[H⁺]ⁿ

The rate for the reaction in trial 2 is:

rate 2 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ

For the reaction in trial 1:

rate 1 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ

If we divide both expressions, we can obtain "b": rate2 / rate1:

rate2/rate1 = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ / k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ

1 = 2ᵇ

b = 0

If we now take the expressions from trial 3 and 1 and divide them, we can obtain "n":

rate 3/rate 1 = k(0.3)ᵃ(0.050)⁰(0.01)ⁿ/ k(0.3)ᵃ(0.050)⁰(0.050)ⁿ

2.1 = 2ⁿ Applying ln to both side of the equation:

ln 2.1 = n ln2

ln2.1/ln2 = n

1 ≅ n

Taking now the reaction in trial 5 and 1 and dividing them:

rate 5/rate 1 = k(0.4)ᵃ(0.050)⁰(0.050) / k(0.3)ᵃ(0.050)⁰(0.050)

4/3 = 4/3ᵃ

a = 1

a)Then the rate law can be written as follows:

rate = k[Acetone][Br₂]⁰[H⁺]

It might be suprising that the rate of bromination of acetone does not depend on the concentration of Br₂. However, looking at the reaction mechanism, you can find out why.

b) Now, we can find the constant k for every experiment and calculate its average value:

rate / [Acetone][Br₂]⁰[H⁺] = k

For reaction 1:

k1 = 5.7 ×10⁻⁵M/s / (0.3 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Reaction 2: k2 = 5.7 ×10⁻⁵M/s / (0.30 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Reaction 3: k3 = 1.2 ×10⁻⁴M/s / (0.30 M)(0.10 M) = 4.0 ×10⁻³ M⁻¹ · s⁻¹

Reaction 4: k4 = 3.1 ×10⁻⁴M/s / (0.40 M)(0.20 M) = 3.9 ×10⁻³ M⁻¹ · s⁻¹

Reaction 5: k5 = 7.6 ×10⁻⁵M/s / (0.4 M)(0.05 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Averge value of k:

k = (k1 + k2 + k3 + k4 + k5)/5 = 3.86 × 10⁻³ M⁻¹ · s⁻¹

3 0

3 years ago

What did Bohr conclude about the atom after observing emission spectrum lines a. Electrons can exist in any energy state but hav

cupoosta [38]

Answer:

Option c: Possible electron energy states are quantized within an atom.

Explanation:

The Bohr's Model of the hydrogen atom consisted of the movements of the electrons around the positively-charged nucleus in circular orbits that have a certain energy state. The energy of that orbit is given by:

$E_{n} = - \frac{Rhc}{n^{2}}$

<em>Where:</em>

E(n): is the energy of an electron in a particular orbit

R: is the Rydberg constant

h: is the Plank constant

c: is the speed of light

n: is a positive integer which corresponds to the number of the orbit

The ground state energy of a electron in the hydrogen atom is equal to -13,6 eV.

Bohr's Model aims to propose that the electron is restrictedly to occupy a certain region in the atom.

Therefore, the conclusion of Bohr after observing emission spectrum lines is that "possible electron energy states are quantized within an atom", so the correct option is c.

I hope it helps you!

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3 years ago

Other than bacteria, which factor leads to nitrogen fixation?

alina1380 [7]

Lightning rain can lead to nitrogen fixation other than bacteria

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A sample of water has a volume of 24.0 millimeters and a mass of 23.8 grams. what is the density

victus00 [196]

24.0mm^3=24÷10÷10÷10cm^3
density=Mass÷Volume

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What was Thompson’s model of the atom called

shtirl [24]

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