<u>Answer:</u> The empirical and molecular formula of the compound is
and
respectively
<u>Explanation:</u>
We are given:
Mass of C = 3.758 g
Mass of H = 0.316 g
Mass of O = 1.251 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.
For Carbon = 
For Hydrogen = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 4 : 4 : 1
The empirical formula for the given compound is 
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:

We are given:
Mass of molecular formula = 130 g/mol
Mass of empirical formula = 68 g/mol
Putting values in above equation, we get:

Multiplying this valency by the subscript of every element of empirical formula, we get:

Hence, the empirical and molecular formula of the compound is
and
respectively
That would be a depression..
Answer:
Acid-base indicators are generally weak proteolytic that change color in solution according to the pH. The acid-base equilibrium of a weak acid type of indicator (HI) in water can be represented as. [I] The acid, HI, and the conjugate base, I−, have different colors. The equilibrium expression for this process is.
<h3>Answer:</h3>
0.64 Moles of Propane
<h3>Explanation:</h3>
Data:
Moles of Carbon = 1.5 mol
Conversion factor = 7 mol C produces = 3 mol of Propane
Solution:
As we know,
7 moles of Carbon produces = 3 moles of Propane
Then,
1.5 moles of Carbon will produce = X moles of Propane
Solving for X,
X = (1.5 moles × 3 moles) ÷ 7 moles
X = 0.6428571 moles of Propane
Or rounded to two significant figures,
X = 0.64 Moles of Propane
Answer: The gas generated by two antacid tablets has a smaller volume.
Explanation:
Since the antiacid is the limiting reagent, we know that the more tablets there are, the more gas there will be.
This means that there will be more gas generated by the four antiacid tablets when compared to the two antiacid tablets, which gives us that the gas generated by the two antiacid tablets has a smaller volume.