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PIT_PIT [208]
2 years ago
7

How many moles of KCl will be produced from 15.0 g of KClO3?

Chemistry
2 answers:
slega [8]2 years ago
7 0

Answer:

0.12 mol KCl

Explanation:

2 KClO3 (s)   2 KCl (s) + 3 O2 (g)

15 g                x mol

x g KCl = 15 g KClO3 x[ (1 mol KClO3)/ (122.5 g KClO3) ] x [(2 mol KCl)/ (2 mol KClO3)]

x g KCl = 0.12 mol KCl

MAXImum [283]2 years ago
7 0

Answer:

0.1224 moles

Explanation:

this is a decompostion reaction

Equation of reaction

2KClO₃ → 2KCl + 3O₂

molar mass of KClO₃ =

molar mass of K = 39.09g/mol

Molar mass of Cl = 35.5g/mol

Molar mass of O = 15.99g/mol

Molar mass of KClO3 = [39 + 35.5 + (3*15.99)]

Molar mass of KClO3 = 122.47g/mol

From the equation of reaction,

2 moles of KClO₃ produces 2 moles of KClO₃

But 1 mole = molar mass

1 mole of KClO₃= 122.47g

2 moles of KClO₃ = 2 * 122.47 = 244.94g

If 244.94g of KClO₃ produces 2 moles of KCl

15 g of KClO₃ will produce x mole of KCl

x = (15 * 2) / 244.94

x = 0.1224moles

15g of KClO₃ will produce 0.1224moles of KCl

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Why is it important to have regular supervision of the weight and measurements in the market
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Answer:

Supervision of weights and measures promotes accurate measurements of goods and services to ensure that everybody gets a fair trade in the marketplace. Not so coincidentally it also is a deterrent to ensure that traders are being honest in their trade practises.

Explanation:

4 0
2 years ago
In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0
2 years ago
Which statement is TRUE regarding the macroscopic and
damaskus [11]

Answer:

Chemists make observations on the macroscopic a scale that lead to conclusions about microscopic features

Explanation:

Many important chemical observations are made on the macroscopic scale. This is because, many of the scientific equipments available are not presently able to provide direct evidence about microscopic processes. Evidences obtained from macroscopic observations could serve as important insights into the nature of certain microscopic processes.

This is evident in the study of the structure of the atom. Most of the evidences that led to the deduction of the atomic structure were obtained from macroscopic evidence but ultimately provided important information about the microscopic structure of the atom.

7 0
2 years ago
Differences between allotropy and isotopy​
raketka [301]

Answer:

Property of an element by virtue of which it exists in two or more forms which differ only in their physical properties is known as allotropy. Allotropes are the different physical forms in which the element can exist. Allotropes are different physical forms of the same element.

Also-

Allotropes are different forms of the same element in the molecular level. Isotopes are different forms of atoms of the same chemical element. The key difference between allotropes and isotopes is that allotropes are considered at the molecular level, whereas isotopes are considered at the atomic leve

Explanation:

~Hope this helps~

3 0
1 year ago
According to the Bohr model, the energy of the hydrogen atom is given by the equation: E = (-21.7 x 10 -19 J)/ n 2 Calculate the
Anit [1.1K]

Answer:

91.6 nm

Explanation:

The energy of the hydrogen atom can be calculated by the emission of a photon. When an electron is excited it goes from to the next energetic level, and when it returns to its ground state, it emits a photon. Hydrogen has only one electron, which is at the level n = 1. So, the equation is given:

E = (-21.7x10⁻¹⁹J)/1²

E = -21.7x10⁻¹⁹J

The energy of the photon is the energy absorbed, and because of that is positive (the opposite of the energy released by the electron). This energy can be calculated by:

E = h*c/λ

Where h is the Planck's constant (6.626x10⁻³⁴ J.s), c is the speed of the light (3.00x10⁸ m/s), and λ is the wavelength of the photon.

21.7x10⁻¹⁹ = 6.626x10⁻³⁴ * 3.00x10⁸/λ

λ = 9.16x10⁻⁸ m

λ = 91.6 nm

7 0
2 years ago
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