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denis23 [38]
3 years ago
7

How many orbitals are completely filled in an atom that has 12 electrons?

Chemistry
1 answer:
Irina-Kira [14]3 years ago
4 0
There are 2 full orbitals and 2 electrons left over.
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The overall fusion reaction is the conversion of Hydrogen to Helium
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How many orbitals are contained in the third main energy level and in the fifth energy level
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There are 9 orbitals in the third energy level and 25 orbitals in the fifth energy level.

I hope this helps you.

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CHEMISTRY HELP PLEASE!
WINSTONCH [101]
1) is called 3-methyl hexane because we choose the longest possible continuous chain which has 6 carbons and start numbering from the side that gives the branch lowest possible number.

2) is called 2-methyl-2-butene because we have to give the double bond lowest possible number but in this case double bond in position 2 from both sides so we start from the side gives the branch number 2 not 3

3)  is called 7-Ethyl-4-decyne because we have to start from the side that gives the triple bond lowest possible number which is 4 and the branch will be at position 7 (note that the name of 10 carbon is incorrectly written in the choices is called decane (as alkane) or decyne when contains triple bond)

4) is called 2,3-Dimethyl pentane because it is 5 carbons (pentane) and we have two branches of the same alkyl (dimethyl) in positions 2 and 3

5) is called 1-Butanol because it contains 4 carbons and has one OH as functional group which take the suffix -ol in position 1 so we said the name as 1-butanol (remember to give the functional group lowest possible number)

6) is called propyl butyl ether because the longest chain is 4 carbons which called butyl and the smallest chain is propyl, it also has another name 1-Propoxy butane <span />
7 0
3 years ago
Read 2 more answers
A compound has a pka of 7.4. to 100 ml of a 1.0 m solution of this compound at ph 8.0 is added 30 ml of 1.0 m hydrochloric acid.
Monica [59]
 Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
6 0
3 years ago
Read 2 more answers
Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25
Veseljchak [2.6K]

Answer: The molecular formula will be C_6H_6O_6

Explanation:

Mass of CO_2 = 17.95 g

Mass of H_2O= 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, =\frac{12}{44}\times 17.95=4.89g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, =\frac{2}{18}\times 4.87=0.541g of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H =  0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.407}{0.407}=1

For H =\frac{0.541}{0.407}=1

For O = \frac{0.410}{0.407}=1

The ratio of C : H : O = 1: 1  : 1

Hence the empirical formula is CHO.

Hence the empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = \frac{44.0}{0.25}\times 1=176g

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6

The molecular formula will be=6\times CHO=C_6H_6O_6

5 0
3 years ago
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