The 2 in front of 
in 
is <u>Coefficient.</u>
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The 
after 
in the same equation is <u>sub </u><u>script</u><u>.</u>
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There are two atoms of in 
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Traditionally they include boron from group 3A, silicon and germanium in group 4A, aresnic and antimony in group 5A and tellurium from group 6A, although sometimes selenium, astatine, polonium and even bismuth have also been considered as metalloids. Typically metalloids are brittle and show a semi-metallic luster.
The six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. Five elements are less frequently so classified: carbon, aluminium, selenium, polonium, and astatine.
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
Yes because the protons are the same but the neutrons change
<span>Water molecules form a complex with metal ions (usually a 6-coordinate complex). And the high charge density on a metal ion draws electrons away from the water molecules, making the O-H bonds more polar than normal. This allows the dissociation of the protons, making solutions of most metal ions acidic</span>
