Idk but look it up on google
Answer:
See explanation
Explanation:
The reactivity of metals has a lot to do with their position in the electrochemical series. However, it is also known that metallic character decreases across the period. This implies that as we move from left to right along the periodic table. Sodium, magnesium, aluminum and silicon continues to decrease in metallic character. As a matter of fact, silicon is a metalloid and not a pure metal.
Sodium reacts with cold water to give a vigorous reaction,magnesium and aluminium reacts with steam at red heat.
Silicon does not react with water, even as steam, under normal conditions.
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
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NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
1)Identify the atoms that are participating in a covalent bond.
2)Draw each atom by using its element symbol. The number of valence electrons is shown by placing up to two dots on each side of the element symbol, with each dot representing a single valence electron.
3)Predict the number of covalent bonds each atom will make using the octet rule.
4)Draw the bonding atoms next to each other, showing a single covalent bond as either a pair of dots or a line representing a shared valence electron pair. If the molecule forms a double or triple bond, use two or three lines to represent the shared electron pairs, respectively.