Explanation:
A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.
Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.
Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.
And, you should ensure that no one else is using an open flame near your experiment.
Thus, we can conclude that following statements are correct:
- You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
- You should ensure that no one else is using an open flame near your experiment.
Answer:
0.019 moles of M2CO3
Explanation:
M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)
From the equation above;
1 mol of M2CO3 reacts to produce 1 mol of BaCO3
Mass of BaCO3 formed = 3.7g
Molar mass of BaCO3 = 197.34g/mol
Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol
Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,
1 = 1
x = 0.019
x = 0.019 moles of M2CO3
Answer:
C
green traveled les distance but still ended up in the same location as red
Answer:
D
Explanation:
We can use the mole ratio to calculate the partial pressure. The total number of moles is 0.2 + 0.2 + 0.1 = 0.5 moles
Now, we know that the mole fraction of the argon gas would be 0.2/0.5
The partial pressure is as follows. To calculate this, we simple multiply the number of moles by the total pressure.
0.2/0.5 * 5 = 1.0/0.5 = 2.00atm
D
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!
