Answer:47.05% is the percent yield of nitrogen in the reaction.
Explanation:
heoretical yield of nitrogen gas = x
Moles of ammonia =
According to reaction,2 moles of ammonia gives 1 mol of nitrogen gas.
Then 2.3529 mol of ammonia will give:
of nitrogen gas
Mass of 1.1764 moles of nitrogen gas,x = 1.1764 mol × 28 g/mol=32.94 g
Experiential yield of nitrogen gas = 15.5 g
Percentage yield:
hope that help
47.05% is the percent yield of nitrogen in the reaction.
Human organisms, and animals can also be infected by Ebola
Answer:
3,6,6...................................
Answer:
A
Explanation:
I guess it's a, because nuclear decay is likely to occur when either the mass or atomic number is greater than 83.
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15
