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MissTica
2 years ago
6

A single electron configuration can represent multiple atoms. The one below represents the following atoms with different charge

s. Determine what charge they need to match the electron configuration. 1s²2s²2p⁶3s²3p⁶4s²3d⁷ ​
Chemistry
1 answer:
valentinak56 [21]2 years ago
6 0

Answer:

To have the electronic configuration equal to 1s²2s²2p⁶3s²3p⁶4s²3d⁷, the chemical element must have an electrical charge equal to 27, that is, it must have 27 electrons, such as Cobalt (Co), for example.

Explanation:

The electronic configuration shown in the question above is known as the Linus Pauling distribution and represents the energy sub-levels that an electrically charged atom can have in relation to the amount of electrons it has.

The layers sub-levels are presented in the following order 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹º 4p⁶ 5s² 4d¹º 5p⁶ 6s² 4f14 5d¹º 6p⁶ 7s² 5f14 6d¹º 7p⁶. Where the small numbers represent the number of electrons in each sub-level and the large numbers represent the layers of electronic distribution.

Accordingly, we can see that an atom that has the configuration 1s²2s²2p⁶3s²3p⁶4s²3d⁷ has 27 electrons, like Cobalt.

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<u><em>I guess this helped you</em></u>

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Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 13. g of butane is m
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<u>Answer:</u> The maximum amount of water that could be produced by the chemical reaction is 20.16 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For butane:</u>

Given mass of butane = 13 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

\text{Moles of butane}=\frac{13g}{58.12g/mol}=0.224mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 70.9 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{70.9g}{32g/mol}=2.216mol

The chemical equation for the reaction of butane and oxygen gas follows:

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

By Stoichiometry of the reaction:

2 moles of butane reacts with 13 moles of oxygen gas

So, 0.224 moles of butane will react with = \frac{13}{2}\times 0.224=1.456mol of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, butane is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water

So, 0.224 moles of butane will produce = \frac{10}{2}\times 0.224=1.12moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.12 moles

Putting values in equation 1, we get:

1.12mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.12mol\times 18g/mol)=20.16g

Hence, the maximum amount of water that could be produced by the chemical reaction is 20.16 grams

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3 years ago
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Answer:

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Explanation:

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n =  \frac{N}{L}  \\

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N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{7.25 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{7.25}{6.02}  \\  = 1.2043189368...

We have the final answer as

<h3>1.20 moles</h3>

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