Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
</span>Look up the freezing point of benzene, then delta T = f.p.=3.5 = Kf*m
<span>Solve for molality. </span>
<span>molality = moles/kg solvent </span>
<span>Solve for moles. </span>
<span>moles acetic acid = grams acetic acid/molar mass acetic acid </span>
<span>Solve for molar mass. You would expect to find 60 for the molar mass CH3COOH.</span>
Answer:
d. End product is that product with a ketone and carboxylic acid.
Explanation:
![{ \sf{NaBH_{4} : }}](https://tex.z-dn.net/?f=%7B%20%5Csf%7BNaBH_%7B4%7D%20%20%3A%20%7D%7D)
Sodium borohydride is a reducing agent, it reduces the ketone to a primary alcohol.
![{ \sf{H _{2} O \: and \: H {}^{ + } }}](https://tex.z-dn.net/?f=%7B%20%5Csf%7BH%20_%7B2%7D%20O%20%5C%3A%20%20and%20%5C%3A%20H%20%7B%7D%5E%7B%20%2B%20%7D%20%7D%7D)
Then acidified water is an oxidising mixture which reverses the reduction reaction.
After the body digest your proteins, they will become your amino acid
"I" symbol means the current goes through the system (imagine the 'I' being a line, like a circuit connecting [power to the device]) "O" symbol means the current does not go through the system. ( the circle is an open circuit, having no power flowing through it