Answer:
a. the mole fraction of CO in the mixture of CO and O2.
mole fraction = moles of CO/ Total moles of the mixture
Mole fraction of CO = 10/(10+12.5)=0.444
b. Reaction - CO(g)+½O2(g)→CO2(g)
Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2
So given,
At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.
3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2
This means that unused mols are : 7mols of CO and 11mols of O2
Total product mixture = 3 + 7 + 11 = 21mols
mole fraction of CO = 7/21 = 0.33
<span>A similar thing occurred with the circles as did with the electroscope. When we initially brought the charged pole close to, the bar pulled in the circle since it was of polarization. At that point, once the bar touched the circle, the pole repulsed the circle. This is on the grounds that once the pole and circle touched, the electrons exchanged thus did the protons, consequently leaving the circle with a positive net charge. The nearer the bar is to the circle the more it repulsed, however, it didn't influence the charge of the circle once the circle was touched by the pole.</span>
Answer: Potassium Iodide, KI
Explanation:
Flame test colors:
Li+ = Crimson Red
Na+ = Bright Orange-Yellow
K+ = Lilac
Addition of nitric acid and silver nitrate (HNO3 and AgNO3),
Cl- = White precipitate
Br- = Creamy precipitate
I- = Yellow Precipitate
Hope this helps, brainliest would be appreciated :)
