Answer:
Yes, the investigations will reach similar conclusions about the reactivity of H2 and Cl2
Explanation:
1. The law of multiple proportions says that when elements form compounds, the proportions of the elements in those chemical compounds can be expressed in small whole number ratios. This means that regardless of whether 1000 times more of the products are used, the reactivity of the products is established by the chemical reaction
2. The law of multiple proportions is an extension of the law of definite composition, which states that compounds will consist of defined ratios of elements.
3. A reaction with more reactants will need more care because more products are produced, which can be toxic
4. H2 and Cl2 reactivity does not depend on the quantities but the chemical properties of each compound
Chemical properties of an atom are based upon the arrangement of valence electrons (electrons which can be gained, lost, or shared).
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Pure Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Br and Br,
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar/Pure Covalent)
For N and O,
E.N of Oxygen = 3.44
E.N of Nitrogen = 3.04
________
E.N Difference
0.40 (Non Polar/Pure Covalent)
For P and H,
E.N of Hydrogen = 2.20
E.N of Phosphorous = 2.19
________
E.N Difference 0.01 (Non Polar/Pure Covalent)
For K and O,
E.N of Oxygen = 3.44
E.N of Potassium = 0.82
________
E.N Difference 2.62 (Ionic)
Answer:
Sand (SiO2 silica) In its pure form it exists as a polymer, (SiO2)n.
Soda ash (sodium carbonate Na2CO3) ...
Limestone (calcium carbonate or CaCo3) or dolomite (MgCO3)
Explanation:
I am pretty sure you know what sand is, Soda ash is a fine white colored powder that is extracted from the ashes of plants growing in sodium rich soils. Limestone is is a common type of carbonate sedimentary rock.
2H₂₍g₎ + O₂ ₍g₎→ 2H₂O
138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O
128 mol H₂O
