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Morgarella [4.7K]
2 years ago
9

Help please!

Chemistry
2 answers:
Mashutka [201]2 years ago
5 0

The electronic configuration of Lv element is 5f^{14} 6d^{10} 7s^2 7p^4.

The electronic configuration of Lv^{+4} is 5f^{14} 6d^{10} 7s^2 .

<h3>What is electronic configuration?</h3>

Electronic configuration, also called electronic structure, is the arrangement of electrons in energy levels around an atomic nucleus.

And also the possible union with hydrogen can give two possible compounds:

LvH_2 \;or LvH_4

This is because the Livermorio compound can have two different oxidation states with the values 2 and 4. When reacting with the hydrogen that its charge is +1, these would be the two possible equations of a client to the different oxidation states with the ones that might react Lv.

It is a synthetic chemical compound and its number in the periodic table is 116.

Learn more about electronic configuration here:

brainly.com/question/13497372

#SPJ1

Akimi4 [234]2 years ago
3 0

#1

Atomic no=116

Lies ahead of Renedium -86

Electronic configuration

  • [Rn]5f¹⁴6d¹⁰7s²7p⁴

#2

Lv^4+ means 4 electrons donated

So it gets removed from p orbital

[Rn]5f¹⁴6d¹⁰7s²

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Answer:

See explaination

Explanation:

Since X is more reactive than Y

=> X is oxidized to X2+ and Y2+ is reduced to Y

Overall cell reaction is:

X(s) + Y2+(aq) => X2+(aq) + Y(s)

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Which of the following is an element?
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Og is the noble gas after Rn. To go from [Rn] to [Og], you must fill four subshells (s, p, d, and f) with a total of 32 electron
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Answer:

See explanation

Explanation:

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3 0
3 years ago
Carbon has four valence electrons. true false
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The answer is true. One neutral atom of carbon has four valence electrons.
 
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Hope I was able to help!
3 0
3 years ago
Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the c
LuckyWell [14K]

Answer:

The percent composition of the compound is 90.5 % C and 9.5 % H

Explanation:

Step 1: Data given

Mass of compound = 9.394 mg

Mass  of CO2 yielded = 31.154 mg

Mass of H2O yielded = 7.977 mg

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2

Step 3: Calculate moles C

moles of C = moles of CO2 * (1 mol C / 1 mol CO2)

moles of C = 7.08 * 10^-4 mol

Step 4: Calculate moles H2O

moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O

Step 5: Calculate moles of H

moles of H = moles of H2O * (2 mol H / 1 mol H2O)

moles of H =  4.43* 10^-4 *2 = 8.86 * 10^-4 mol H

Step 6: Calculate mass of C

mass C = moles C * molar mass C

mass C = 7.08 * 10^-4 mol*12.01 g/mol

mass C = 0.0085 grams

Step 7: Calculate mass of H

mass H = moles H * molar mass H

mass H = 8.86 * 10^-4 mol*1.01 g/mol

mass H = 0.000894 grams

Step 8: Calculate total mass of compound =

0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg

Step 9: Calculate the percent composition:  

% C = (8.50 mg / 9.394 mg) x 100 = 90.5%  

% H = (0.894 mg / 9.394 mg) x 100 = 9.5%

The percent composition of the compound is 90.5 % C and 9.5 % H

6 0
4 years ago
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