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
Answer:
atom
Explanation:
look on the periodic table
Answer:
0.1077 grams
Explanation:
First we will employ the ideal gas law to determine the number of moles of nitrogen gas.
PV=nRT
P=2 atm
V=20L
R=0.08206*L*atm*mol^-1*K^-1
T=323.15 K
Thus, 2atm*20L=n*0.08206*L*atm*mol^-1*K^-1*323.15K
K, atm, and L cancels out. Thus n=2*20mol/0.08206*323.15=1.5 moles
Lastly, we must convert the number of moles to grams. This can be done by dividing the number of moles by the molar mass of nitrogen gas, which is 14 grams.
1.5/14=0.1077 grams
12.) 15 : 10 Times Each One By 0.3
15*0.3= 4.5
15-4.5 = 10.5
10*0.3= 3
10-3=7
The Scale Factor Is 10.5:7
