Explanation:
We have to find the number of molecules that are present in 0.54 g of Ca(NO₃)₂.
First we have to convert the mass of our sample into moles of Ca(NO₃)₂. We will use the molar mass of Ca(NO₃)₂ to do that.
molar mass of Ca = 40.08 g/mol
molar mass of N = 14.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of Ca(NO₃)₂ = 1 * 40.08 g/mol + 2 * 14.01 g/mol + 6 * 16.00 g/mol
molar mass of Ca(NO₃)₂ = 164.10 g/mol
mass of Ca(NO₃)₂ = 0.54 g
moles of Ca(NO₃)₂ = 0.54 g * 1 mol/(164.10 g)
moles of Ca(NO₃)₂ = 0.00329 moles
According to Avogadro's number there are 6.022 *10^23 molecules in 1 mol of molecules. We can use that relationship to find the number of molecules that are present in our sample.
6.022 *10^23 molecules = 1 mol
molecules of Ca(NO₃)₂ = 0.00329 moles * 6.022 *10^23 molecules/(1 mol)
molecules of Ca(NO₃)₂ = 2.0 * 10^21 molecules
Answer: there are 2.0 * 10^21 molecules of Ca(NO₃)₂ in 0.54 g of it.
Answer:
Mole ratio for a compound
The chemical formula tells us the mole ratio.
CO2 = 1 CO2 molecule : 1 C atom : 2 O atoms.
Mole ratio for a reaction
The balanced chemical reactions tells us.
C12H22O11 + 12 O2 12 CO2 + 11 H2O
1 C12H22O11 molecule: 12 O2 molecules : 12 CO2
molecules : 11 H2O molecules.
Applications of the mole ratio concept
grams <--> moles <--> moles <--> grams
Explanation:
Answer:
See figure 1
Explanation:
For this question, we have to remember that in the lewis structures all atoms must have<u> 8 electrons</u>. And each atom would have a different value of <u>valence electrons</u>:
Carbon => 4
Oxygen=> 6
Hydrogen=> 1
Additionally, for the <u>hybridizations</u> we have to remember that:
=> 4 single bonds
=> 1 double bond
=> 1 double bond
With this in mind, the formaldehyde and formic acid would have
carbons and the ethanol an
carbon.
Finally, for the oxidation state. We have to remember that <u>if we have more bonds with oxygen, we will have more oxidation</u>. Therefore, the carbon that has more oxidation is the one in the formic acid (we have several bonds with oxygen).
See figure 1
I hope it helps!