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alex41 [277]
3 years ago
6

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

Chemistry
1 answer:
Blababa [14]3 years ago
6 0

Answer:

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.

Element Atomic number Electron configuration

sulfur 16 1s22s22p63s23p4

chlorine 17 1s22s22p63s23p5

argon 18 1s22s22p63s23p6

or the

Two electrons

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

Explanation:

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Answer: 1. 2.7 moles  of ammonia are formed

2. 12.0 moles  of hydrogen are required

3. 2.0 moles of nitrogen are required

Explanation:

The balanced chemical equation is:

N_2+3H_2\rightarrow 2NH_3

According to stoichiometry:

3 moles of hydrogen form = 2 moles of ammonia

Thus 4.0 moles of hydrogen form =\frac{2}{3}\times 4.0=2.7moles of ammonia

According to stoichiometry:

2 moles of ammonia are formed by = 3 moles of hydrogen

Thus 8.0 moles of ammonia are formed by  =\frac{3}{2}\times 8.0=12.0moles of hydrogen

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The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon
tigry1 [53]

Answer:

\boxed{\text{300 g}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:         32                          60

           CH₃OH + CO ⟶ CH₃COOH

m/g:        160

(a) Moles of CH₃OH

\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}

(b) Moles of CH₃COOH

\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}

(c) Mass of CH₃COOH

\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}

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