Ask question

Ask question

All categories

3 years ago

8

I have a balanced equation. I can convert moles of one substance to moles of another substance using the _____ ratio as a unit f

actor

1 answer:

Olin [163]3 years ago

7 0

Answer: sorry can't help ya

Explanation:

Sorry it's just I don't really know how to answer this question myself wish you the best of luck ;3

You might be interested in

What is the meaning of the term polar as applied to chemical bonding?

kicyunya [14]

Polar<span> covalent </span>bonding<span> is a type of </span>chemical bond <span>where a pair of electrons is unequally shared between two atoms.</span>

8 0

3 years ago

Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...

g100num [7]

Answer:

O₂; KCl; 33.3

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

2KCl + 3O₂ ⟶ 2KClO₃

n/mol: 100.0 100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

$\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}$

(ii) From O₂

$\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}$

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

$\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}$

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

3 0

3 years ago

Relate what you have learned about the word quantum to the Bohr model.

ss7ja [257]

I forgot what quantum means to be honest, the Bohr model In atomic physics, the Bohr model or Rutherford–Bohr model, presented by Niels Bohr and Ernest Rutherford in 1913, is a system consisting of a small, dense nucleus surrounded by orbiting electrons—similar to the structure of the Solar System, but with attraction provided by electrostatic forces in place of gravity. After the cubical model (1902), the plum pudding model (1904), the Saturnian model (1904), and the Rutherford model (1911) came the Rutherford–Bohr model or just Bohr model for short (1913). The improvement over the 1911 Rutherford model mainly concerned the new quantum physical interpretation.

7 0

3 years ago

A particle that orbits the nucleus in an atom is called a(n

Morgarella [4.7K]

Particles that orbit the nucleus are called electrons.

Explanation: Electrons are negatively charged particles arranged in orbits around the nucleus of an atom and determining all of the atom's physical and chemical properties except mass and radioactivity.

7 0

3 years ago

Read 2 more answers

1. A chemist prepares hydrogen fluoride by means of the following reaction:

Natasha_Volkova [10]

Answer:

a) <em>Theoretical Yield of HF = 5.64 grams</em>

b) <em>Percentage Yield = 39%</em>

Explanation:

Reaction Given:

CaF2 + H2SO4 -> CaSO4 + 2HF

CaF2 = 11g

H2SO4 = Used in excess

HF = 2.2 g production = Actual Yield

So, Let's write down the molar masses:

Molar Mass of CaF2 = 78 g /mol

Molar Mass of HF = 20 g/mol

From the reaction, we can see the 1 mole of CaF2 gives the 2 moles of HF

i.e

a) Theoretical Yield of HF:

1 mole CaF2 = 2 moles HF

78 g CaF2 = 2 x 20 g of HF

78 g CaF2 = 40 g of HF

1 g CaF2 = 40g/78g of HF

And in the question it is given that chemist used 11 g of CaF2 so,

1 x 11 g of CaF2 = 11 x 40/78 g of HF

11 g of CaF2 = 440/78 g of HF

11 g of CaF2 = 5.64 g of HF

And this is the theoretical yield

<em>Theoretical Yield of HF = 5.64 grams</em>

b) Now, calculate the Percentage Yield of HF

<em>Percentage Yield = Actual Yield /Theoretical Yield x 100</em>

Percentage Yield = 2.2 g /5.64 g x 100

Percentage Yield = 39%

8 0

3 years ago