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vladimir1956 [14]
3 years ago
15

Which of the following electron configurations correspond to an atom with an electron in an excited state?

Chemistry
1 answer:
ryzh [129]3 years ago
7 0

The correct answer is option (B) [Ne]3s¹3p⁶.

The electron configuration of Cl is [Ne]3s²3p⁵ in the ground state.

The electron configuration of Cl is [Ne]3s¹3p⁶ in the excited state.

In the excited state one electron from the 2s orbital jumps to the 2p orbital. As a result the 2p orbital becomes completely filled with 6 electrons and the 2s orbital is partially filled with only 1 electron. To be completely filled a p and a s orbital can have a maximum of 6 and 2 electrons respectively.

Thus [Ne]3s¹3p⁶ electron configuration correspond to an atom with an electron in an excited state.

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A student titrated a solution containing 3.7066 g of an unknown Diprotic acid to the end point using 28.94 ml of 0.3021 M KOH so
uranmaximum [27]

Answer:

<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>

Explanation:

[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>

Ionic equation:

{ \bf{2OH { }^{ - }  _{(aq)} + 2H { }^{ + } _{(aq)}→  2H _{2} O _{(l)} }}

first, we get moles of potassium hydroxide in 28.94 ml :

{ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

{ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\  = { \underline{423.95 \: g \approx424 \: grams}}

for the molar mass:

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