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maks197457 [2]
2 years ago
15

6. The central selenium atom in selenium hexafluo-

Chemistry
1 answer:
serg [7]2 years ago
5 0

Given what we know, we can confirm that when forming an expanded octet, selenium hexafluo ride will hold 6 pairs of shared electrons around its center selenium atom.

<h3>What is an expanded octet?</h3>
  • An expanded octet is when an atom can hold more than 8 valence electrons in its outer shell.
  • This is possible for those elements in period four of the periodic table.
  • Elements that are capable of this form what we call hypervalent compounds.

Therefore, given that during an expanded octet formation, the central atom is capable of holding more than 8 valence electrons in its outer shell, the central atom of a selenium hexafluo ride compound will have 12 electrons being shared, which results in 6 pairs, making C the correct answer.

To learn more about valence electrons visit:

brainly.com/question/7223122?referrer=searchResults

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The lead pipe has a mass of 200 g, how much will be left after 6 months?
kumpel [21]

Answer:

33g

Explanation:

200 divided by 6=33.33

5 0
3 years ago
1N2 + 3H2 --&gt;
Hunter-Best [27]

Answer:

28.23 g NH₃

Explanation:

The balanced chemical equation is:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Thus, 1 mol of N₂ reacts with 2 moles of H₂ to produce 2 moles of NH₃. We convert the moles to mass (in grams) by using the molecular weight (MW) of each compound:

MW(N₂) = 2 x 14 g/mol = 28 g/mol

mass N₂= 1 mol x 28 g/mol = 28 g

MW(H₂) = 2 x 1 g/mol = 2 g/mol

mass H₂ = 3 mol x 2 g/mol = 6 g

MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

mass NH₃= 2 moles x 17 g/mol = 34 g

Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

5 0
2 years ago
The specific heat of ethanol is 2.44 J/g ֯C. How many kJ of energy are required to heat 50.0 grams of ethanol from -20 ֯C to 68
Aliun [14]

Answer:

Heat energy required (Q) = 10.736 KJ

Explanation:

Given:

Specific heat of ethanol (C) = 2.44 J/g °C

Mass of ethanol (M) = 50 gram

Initial temperature (T1) = -20°C

Final temperature (T1) = 68°C

Find:

Heat energy required (Q) = ?

Computation:

Change in temperature (ΔT) = 68°C - (-20°C)

Change in temperature (ΔT) = 88°C

Heat energy required (Q) = mC(ΔT)

Heat energy required (Q) = (50)(2.44)(88)

Heat energy required (Q) = 10,736 J

Heat energy required (Q) = 10.736 KJ

5 0
3 years ago
Which statement best describes the benefit of having a universal system of measurement
Viefleur [7K]
I think it would be c
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3 years ago
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Roberto examines two molecular models. He sees that the atoms are the same in each model and are attached in the same order in e
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Nao me denuncia eu so quero pontos
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