Answer:
Explanation:Remember that c is always about 300,000 km/s. Use the formula you identified to explain how the two factors in the formula are related on the electromagnetic spectrum.
<span>When you have a polar molecule, your bonds will not cancel out. This means that in a polar bond, the electronegativity of the atoms will attend to be different. For non-polar bonds the electro-negativity of the atoms will also be equal. In a polar bond you will have an unequal sharing of electron pairs which causes a molecular dipole. I hope this helps.
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The molar mass and the mass of the substance give the moles of the substance. The 10-gram sample of neon (Ne) has the fewest atoms. Thus, option b is correct.
<h3>What are moles?</h3>
Moles of the substance are the ratio of the molar mass and the mass of the substance. The moles of each atom with respect to the Avagadro's number (Nₐ) is given as,
Moles = mass ÷ molar mass
- Moles of carbon (C) = 10 ÷ 12
= 0.83 Nₐ atoms
- Moles of neon (Ne) = 10 ÷ 20
= 0.5 Nₐ atoms
- Moles of fluorine (F) = 10 ÷ 19
= 0.52 Nₐ atoms
- Moles of nitrogen (N) = 10 ÷ 14
= 0.7 Nₐ atoms
- Moles of oxygen (O) = 10 ÷ 16
= 0.63 Nₐ atoms
Therefore, option b. 10 gm neon has the fewest atoms.
Learn more about moles here:
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Example #1 write the chemical formula for Calcium Oxide
Step 1
-Find the Atomic symbol of the metal and non-metal on the periodic table
CaO
Step 2
Find the charges for Calcium and Oxygen
which are Ca 2+ and O 2-
Step 3
Balance out the charges
They are already balanced out
Here´s the chemical formula for Calcium Oxide
CaO
Example 2 write the chemical formula for Aluminum Oxide
Step 1
-Find the Atomic symbol of the metal and non-metal on the periodic table
A
l
O
Step 2
Find the charges for Aluminum and Oxygen
Which are Al 3+ and O 2-
Step 3
Balance out the charges
You need 2 Aluminum and 3 Oxygen to balance the charges
Al 3+ O 2-
Al 3+ O 2-
= 6 + O 2-
= 6-
Step 4
If you need more then one element to balance out the charges you identify that by using subscripts
Heres your chemical formula for Aluminum Oxide
