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3 years ago

Why alkali metals easly ionize

1 answer:

7 0

Here’s what I found:

It takes very little energy to remove that outermost electron from an alkali metal. Thus, alkali metals easily lose their outermost electron to become a +1 ion. ... In fact, as you go down the 1A column, the first ionization energies get lower and lower, making cesium the most easily ionized element on the periodic table.

So basically it’s because part of what makes alkali metals so reactive is that they have one electron in their outermost electron layer.

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Ice cubes with salt melt faster. When you add salt, it dissolves in the water of the ice cube. The freezing temperature of salt water is 32 degrees Fahrenheit lower than that of fresh water.

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3 years ago

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Answer:

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3 years ago

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A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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3 years ago

7. What happen if we don't have water?

Ahat [919]

Answer:

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3 years ago

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At a particular temperature, 12.0 moles of so3 is placed into a 3.0-l rigid container, and the so3 dissociates by the reaction 2

saul85 [17]

2 SO₃ --> 2 SO₂ + O₂
I 12 0 0
C -2x +2x +x
---------------------------------------------
E 12-2x 2x x

Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5

The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]²
where the unit used is conc in mol/L.

K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
<em>K = 0.0556</em>

4 0

3 years ago