Answer:
2 C2H5OH+ 21 o2 —> 4 CO2+ 6 H2o6
Answer:
Amount of sugar in solution = 82.5 gram
Explanation:
Given:
Sugar in solution percentage = 15%
Total solution = 550 grams
Find:
Amount of sugar in solution
Computation:
Amount of sugar in solution = Sugar in solution percentage x Total solution
Amount of sugar in solution = 15% X 550
Amount of sugar in solution = 82.5 gram
If matter was not composed of atoms, the law definite proportions and law of multiple proportions will not hold true because compound would have random composition.
<h3>What is the law of definite proportions and law of multiple proportions?</h3>
The law of definite proportions states that all pure samples of a compound contains the same element in a fixed mass ratio.
The law of multiple proportions states that if an element combines with another element to form more than one compound, the mass of the element combines with a fixed mass of the other element in whole number multiples ratios.
Assuming matter was not composed of atoms, the law definite proportions and law of multiple proportions will not hold true because, their will be no definite ratio or composition in compounds.
The law definite proportions and law of multiple proportions are required for the atomic theory as they help prove that matter consists of definite unique particles called atoms.
In conclusion, the existence of atoms is proved true by the law definite proportions and law of multiple proportions.
Learn more about law of multiple proportions at: brainly.com/question/2624012
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Answer:
M = 3.69 M.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the molar concentration of the 1.29 moles of KCl in 350 mL of solution by recalling the mathematical definition of molarity as the division of the moles by the volume in liters, in this case 0.350 L; thus, we proceed as follows:
Which gives molar units, M, or just mol/L.
Regards!
Mass of Oxygen (O₂) : = 88.16 g
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
C₆H₁₄+ O₂ → CO₂ + H₂0
25 g C₆H₁₄
Required
mass of oxygen (O₂)
Solution
Balanced equation
2C₆H₁₄ + 19O₂ ⇒12 CO₂ + 14 H₂O
mol C₆H₁₄ (MW=86,18 g/mol) :
= mass : MW
= 25 g : 86.18 g/mol
= 0.29
From the equation, mol ratio of C₆H₁₄ : O₂ = 2 : 19, so mol O₂ :
= 19/2 x mol C₆H₁₄
= 19/2 x 0.29
= 2.755
Mass O₂(MW=32 g/mol) :
= 2.755 x 32
= 88.16 g
