In accordance with Dalton's Law of multiple proportions
<h3>Further explanation</h3>
Given
6.0g of carbon
22.0g or 14.0g of product
Required
related laws
Solution
the amount of air present ⇒ as an excess or limiting reactant
- air(O₂) as a limiting reactant(product=14 g)
C+0.5O₂⇒CO
6 + 8 = 14 g
mol O₂=8 g : 32 g/mol=0.25
mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)
mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g
- air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant
C+O₂⇒CO₂
6 + 16 = 22 g
mol C = 6 g : 12 g/mol = 0.5
mol O₂ = 16 g : 32 g/mol=0.5
mol CO₂ = 22 g : 44 g/mol = 0.5
if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.
CO = 6 : 8
CO₂ = 6 : 16
the ratio O = 8 : 16 = 1 : 2
In accordance with Dalton's Law of multiple proportions
Orange i believe so or if not blue
Answer:
Valence Electrons
Explanation:
Valence electrons are the electrons in the outermost shell and can be used to form bonds.
Answer
Depends on type of mixture. But I think separating the different sized particles through filtration would be a sufficient answer for middle school level unless they have taught you about other mixtures.
Explanation:
Hey,
There are thousands of way to separate mixtures. Each way is specific to the type of mixture. If the mixture is homogeneous processes like distillation can be employed. For heterogeneous mixture filtration can be used to separate particles of different sizes.
