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Goshia [24]
2 years ago
13

All organic compounds have at least ____ carbon and two ____

Chemistry
1 answer:
denpristay [2]2 years ago
7 0
All organic compounds have at least 1 carbon and 2 hydrogen atoms.
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How many grams are in 8.2 x 1022 molecules of N216?
hoa [83]

Answer:

N2I6 = 789 g

N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles

N2I6=0.136molesx789g/mole=107g=110g

----------------------

Hope this helps, have a BLESSED AND WONDERFUL DAY!

- Cutiepatutie ☺❀❤

7 0
3 years ago
Explain what makes a covalently bonded molecule polar. Use the molecules of CH4 and CH3F as examples in your explanation.
RideAnS [48]

Answer:

See explanation

Explanation:

The magnitude of electronegativity difference between atoms in a bond determines whether that bond will be polar or not.

If the electronegativity difference between atoms in a bond is about 1.7, the bond is ionic. If the electronegativity difference is greater than 0.4 and less than 1.7, the bond will have a polar covalent character. Lastly, if the electronegativity difference between the bond is less than or equal to 0.4, the covalent bond is non polar.

The electronegativity difference between carbon and hydrogen is about 0.4 which corresponds to a nonpolar covalent bond hence the molecule is nonpolar.

The electronegativity difference between carbon and fluorine is about 1.5 indicating a highly polar bond. This gives CH3F an overall dipole moment thereby making the molecule polar.

8 0
2 years ago
___CH4 + ___O2 → ___CO2 + ___H2O
DaniilM [7]
CH4 + 2O2 =======> CO2 + 2H2O
5 0
3 years ago
A sample of a pure compound that weighs 59.8 g contains 27.6 g Sb (antimony) and 32.2 g F (fluorine). What is the percent compos
Rama09 [41]

Answer:

53.85%

Explanation:

Data obtained from the question include:

Mass of antimony (Sb) = 27.6g

Mass of Fluorine (F) = 32.2g

Mass of compound = 59.8g

Percentage composition of fluorine (F) =..?

The percentage composition of fluorine can be obtained as follow:

Percentage composition of fluorine = mass of fluorine/mass of compound x 100

Percentage composition of fluorine = 32.2/59.8 x 100

= 53.85%

Therefore, the percentage composition of fluorine in the compound is 53.85%

3 0
3 years ago
20 Points!!!!!!!!!!!! Please Help!!!!
mamaluj [8]
Number 1: (A.)
Number 2: (A.)
Number 3: (B.)

I'm probably wrong but that is what i think
3 0
3 years ago
Read 2 more answers
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