Answer:
324.18 g/mol
Explanation:
Let the molecular mass of the antimalarial drug, Quinine is x g/mol
According to question,
Nitrogen present in the drug is 8.63% of x
So, mass of nitrogen = 
Also, according to the question,
2 atoms are present in 1 molecule of the drug.
Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)
So, mass of nitrogen = 14.01×2 = 28.02
These 2 must be equal so,

solving for x, we get:
<u>x = 324.18 g/mol</u>
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The answer is 6.1*10^-3 atm.
The pictures and explanations are there.
In a chemical reaction,
the limiting reagent is the chemical being used up while the excess reactant is
the chemical left after the reaction process.
Before calculating the limiting
and excess reactant, it is important to balance the equation first by stoichiometry.
C25N3H30Cl + NaOH = C25N3H30OH + NaCl
Since the reaction is already balanced, we can now identify which
is the limiting and excess reagent.
First, we need to determine the number of moles of each chemical
in the equation. This is crucial for determining the limiting and excess reagent.
<span>Assuming that there is the
same amount of solution X for each reactant</span>
1.0 M NaOH ( X ) = 1.0
moles NaOH
1.00 x 10-5 M C25N3H30Cl
( X ) = 1.00 x 10-5 moles C25N3H30Cl
<span>The result showed that the
crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this
chemical reaction is crystal violet and the excess reactant is NaOH.</span>
2 C2H2 + 5 02 > 4 CO2 + 2 H2O
Products - Reactants ( all units are kJ/mo1):
(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1
-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1
answer: -1255.6 kJ/mo1