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Dimas [21]
3 years ago
10

Plzzzzzzzzzzzzzzzz help

Chemistry
2 answers:
DiKsa [7]3 years ago
6 0

Answer:

The answer is C.

Explanation:

I just know.

Anna [14]3 years ago
5 0

Answer:

Your correct.

Explanation:

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Enter the correct ground-state (or lowest energy) configuration based on the number of electrons: 1s^4 2s^4 2p^12.
Blababa [14]

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The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.

or the

FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.

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sulfur 16 1s22s22p63s23p4

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or the

Two electrons

Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.

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7 0
3 years ago
One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO
zhenek [66]
Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
 2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s)  and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
                                                                                     = 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
                                                         = 1000 x 10³ g / 226 g mol⁻¹
                                                         = 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x \frac{3}{2}
                                                     = 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
                                        = 703540.02 g
                                        = 703.540 kg

Hence, the theoretical yield of Na₂CO₃ =  703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
                            = 92.34%
7 0
3 years ago
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