Why isn't hydrogen considered an alkali metal ?

2 answers:

7 0

Hydrogen is a non-metal, it is one of the two elements on the periodic table that only need two valence electrons to be stable. Hydrogen doesn't share many properties with the other Alkali Metals. It has a much higher electronegativity, doesn't form ionic bonds, doesn't react with the same things that the other elements in group one do (like water.) The only reason hydrogen is place there is because it has one valence electrons like all the other alkali metals and there is not where else to place hydrogen, because it doesn't really fit in with any group.

Maurinko [17]3 years ago

4 0

<span>Hydrogen isn't classified as an alkali metal because it exhibits several properties that are not found in the alkali metal group, including the tendency to be found as a diatomic molecule, a tendency to form weak bonds with other bound hydrogen atoms nearby, weak electromagnetism, a low activity series placement, and the tendency to be found as a gas. Because of all this, it's impossible to classify it with any group, although it does seem to be closest to nonmetals.</span>

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2 years ago

Explain why lithium is a strong reducing agent, whereas fluorine is a strong oxidizing agent.

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3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$