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Alina [70]
3 years ago
8

Which atom in the ground state has a partially filled second electron shell?

Chemistry
2 answers:
adell [148]3 years ago
7 0
The answer is lithium.

This is because it has an electron configuration of 1s2, 2s1 (also written as [He]2s1)
serious [3.7K]3 years ago
4 0
Atoms have electrons filled in energy shells.
1. H - hydrogen atom has one electron in the First energy shell. Therefore hydrogen has a partially filled first energy shell

2.Li - Li electron configuration is 2,1
The outermost energy shell is the second energy shell in which there is only one electron
Therefore the second energy shell is partially filled. This is the correct answer

3. K - electron configuration is 2,8,8,1
The outermost energy shell is the fourth energy shell which is partially filled. The second energy shell is completely filled

4.Na - electron configuration is 2,8,1
The outermost energy shell is the third energy shell which is partially filled
Second energy shell is completely filled

From the given options Li is the only element with a partially filled second energy shell
Answer is Li
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300 gallons

Explanation:

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2 years ago
PLEASE HELP!!! NEED TO PASS THIS TO THE FIRST SEMESTER!
Cloud [144]

Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)

2.CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)

Explanation:

Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)

This is acid - base type reaction where

CH_{3}COOH(aq) act as Acid

NaHCO_{3}(aq) act as weak base

Reactant :CH_{3}COOH(aq) ,NaHCO_{3}

Number of atoms of :

C = 2 (CH_{3}COOH(aq)) + 1 (NaHCO_{3})

   = 2 + 1

   = 3

H  = 4(CH_{3}COOH(aq)) + 1 (NaHCO_{3})

     = 4 + 1

       5

O = 2(CH_{3}COOH(aq)) + 3 (NaHCO_{3})

    = 5

Na = 1 (NaHCO_{3})

     = 1

Product :  CO_{2}(g),H_{2}O(l) , CH_{3}COONa(aq)

Number of atoms :

C = 1(CO_{2}(g)) + 2(CH_{3}COONa(aq))

   = 1 + 2

   = 3

H = 2(H_{2}O(l)) + 3(CH_{3}COONa(aq))

   = 2 + 3

   = 5

O = 1(H_{2}O(l)) + 2(CH_{3}COONa(aq))

        +2(CO_{2}(g)

    = 1 + 2 + 2

    = 5

Na = 1(CH_{3}COONa(aq)

     = 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

2.CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + CO_{2}(g)

This is double displacement reaction .

Check the balancing in both reactant and products should be :

Na = 2

H = 2

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2 years ago
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steposvetlana [31]

Answer:

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Explanation:

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Ymorist [56]

The   molarity  of (HNO₃) that was used  if 2.00 L must  be used   to prepare 4.5 L  of a 0.25M HNO₃ solution is   0.563 M


 <u><em>calculation</em></u>

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   M₁  is therefore = M₂V₂/V₁

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