Which group will have an electron configuration that ends in s2?

Chemistry

1 answer:

Step2247 [10]3 years ago

8 0

<h2>Answer: c . Alkaline earth metals (Group 2)</h2>

<h3>Explanation:</h3>

Group 2 metals have 2 electrons in their outer shell. These two electrons are usually found in the s orbital, hence the s².

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Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f

Korvikt [17]

Answer:

$55.2kgNa_{3}AlF_{6}$

Explanation:

1. First balance the equation for the synthesis of cryolite:

$Al_{2}O_{3}_{(s)}+6NaOH_{(l)}+12HF_{(g)}=2Na_{3}AlF_{6}+9H_{2}O_{(g)}$

2. Find the limiting reagent between the $Al_{2}O_{3},NaOH$ and $HF$

- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:

$13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}$

$55.4kgNaOH*\frac{1molNaOH}{40kgNaOH}*\frac{1000g}{1kg}=1385molesNaOH$ $55.4kgHF*\frac{1molHF}{20kgHF}*\frac{1000g}{1kg}=2770molesHF$

- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:

$Al_{2}O_{3}:\frac{131}{1}=131$

$NaOH:\frac{1385}{6}=231$

$HF:\frac{2770}{12}=231$

The $Al_{2}O_{3}$ is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

$13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}$

3 0

4 years ago

In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?

klio [65]

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the <u>vibration of the bonds</u> by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is <em>a specific energy that generates a specific vibration</em>. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the <u>lower the wavenumber we will have less energy</u>. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have <u>heteroatoms</u> (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of <u>resonance structures</u> which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the <u>cyclohexanone.</u>

See figure 1

I hope it helps!