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Arlecino [84]
2 years ago
10

Explain why the molar ratio is needed to determine the theoretical mass ratios.

Chemistry
1 answer:
choli [55]2 years ago
3 0

Answer:

The overview of the subject is outlined underneath in the summary tab.

Explanation:

  • The molar ratio seems to be essentially a balanced chemical equilibrium coefficient that implies or serves as a conversion factor for the product-related reactants.
  • This ratio just says the reactant proportion which reacts, but not the exact quantity of the reacting product. Consequently, the molar ratio should only be used to provide theoretical instead of just a definite mass ratio.

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Planck suggested that light/energy was absorbed/released in certain amounts, called quanta.
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Which of the following represents the atomic number of an element?
Anna11 [10]

Answer:

The atomic number is the number of protons in the nucleus

4 0
2 years ago
How many moles of CaCO3 are needed to react with 12.5 mol SO2​
grandymaker [24]
<h3><u>Answer;</u></h3>

= 12.5 Moles of CaSO3

<h3><u>Explanation</u>;</h3>

The reaction between CaCO3 and SO2 is given by the equation.

CaCO3(s) + SO2(g) → CaSO3(aq) + CO2(g)

The mole ratio between CaCO3 and SO2 is 1 : 1;

1 mole of CaCO3 reacts with 1 mole SO2 to form CaSO3 and CO2

Therefore;

<em>12.5 moles of SO2 will require 12.5 moles of CaSO3</em>

7 0
3 years ago
2. What is the mass of 5.33 x 10 moles of aluminum hydroxide?​
bearhunter [10]
<h3>Answer:</h3><h3>1865.5g</h3><h3>Explanation:</h3><h3 /><h2> first the chemical formular for ammonium hydroxide is NH4OH</h2><h3>its molarmass is given as N=14H=1O=16 </h3><h3> so we have 14 +1(2) +16+1 =35</h3><h2>also no of moles = mass / molarmass</h2><h3> we have 5.33×10 = mass/35 </h3><h2>therefore mass = 35 ×5.33×10 = 1865.5g</h2>
8 0
2 years ago
Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empi
JulijaS [17]

1. The empirical formula of the hydrocarbon is CH₃

2. The molecular formula of the hydrocarbon is C₂H₆

<h3>How to determine the mass of Carbon </h3>
  • Mass of CO₂ = 1.47 g
  • Molar mass of CO₂ = 44 g/mol
  • Molar of C = 12 g/mol
  • Mass of C =?

Mass of C = (12 / 44) × 1.47

Mass of C = 0.4 g

<h3>How to determine the mass of H</h3>
  • Mass of compound = 0.5 g
  • Mass of C = 0.4 g
  • Mass of H = ?

Mass of H = (mass of compound) – (mass of C)

Mass of H = 0.5 – 0.4

Mass of H =0.1 g

<h3>1. How to determine the empirical formula </h3>
  • C = 0.4 g
  • H = 0.1 g
  • Empirical formula =?

Divide by their molar mass

C = 0.4 / 12 = 0.03

H = 0.1 / 1 = 0.1

Divide by the smallest

C = 0.03 / 0.03 = 1

H = 0.1 / 0.03 = 3

Thus, the empirical formula of the compound is CH₃

<h3>2. How to determine the molecular formula</h3>
  • Empirical formula = CH₃
  • Molar mass = 30 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Learn more about empirical formula:

brainly.com/question/24297883

#SPJ1

4 0
2 years ago
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