Easy stoichiometry conversion :)
So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.
So, our first step would look like this:
10.0
------
1
Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.
So, our 2nd step would look like this:
1 mole CO2
-----------------
84.007g NaHCO3
When we put it together: our complete stoichiometry problem would look like this:
10.0g NaHCO3 1mol CO2
---------------------- x -------------------------
1 84.007g NaHCO3
Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)
And then....
Divide the top answer by the bottom answer.
10.0/84.007 is 0.119
So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.
Hope I could help!
Why did you post this again? Because you're lazy and wanted people to not see my post? Once again, ONE QUESTION AT A TIME. We're here to help you with a tough question or work you through it, not do all your homework.
Answer:
Ground state
the state with the smallest amount of energy.
Answer:
stay the same.
Explanation: Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):
Na: 1s2 2s2 2p6 3s1
Ar: s2 2s2 2p6 3s2 3p6
As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:
Na: 3s1
Ar: 3s2, 3p6
All the new electrons electrons filled the third energy level (3s and 3p). So the energy level does not change, just the orbitals.
Answer:
AuCl
Explanation:
Given parameters:
Mass of Gold = 2.6444g
Mass of Chlorine = 0.476g
Unknown:
Empirical formula = ?
Solution:
Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.
Elements Au Cl
Mass 2.6444 0.476
Molar mass 197 35.5
Number of moles 2.6444/197 0.476/35.5
0.013 0.013
Divide by the
smallest 0.013/0.013 0.013/0.013
1 1
The empirical formula of the compound is AuCl
