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!
Answer:
600K
Explanation:
PV=nRT
T=PV/nR
= 1.6atm* 15.0L/ 0.5mol*0.0821LatmK^-1mol^-1
=600K
The flame goes an Orange-Red colour.
Answer:
electrophile(H⁺) is needed to react with alkene in the first step and nucleophile (OH⁻) is not available in the first step
Explanation:
