Mass of O₂ formed = 7.84 g
Further explanationGiven
Reaction
2KClO₃(s) ⇒2KCl(s) + 3O₂(g)
P water = 23.8 mmHg
P tot = 755 mmHg
V = 6.22 L
T = 25 + 273 = 298 K
Required
mass of O₂
Solution
P tot = P O₂ + P water
P O₂ = P tot - P water
P O₂ = 755 - 23.8
P O₂ = 731.2mmHg = 0.962 atm
Moles O₂ :
Ideal gas law :
n = PV/RT
n = 0.962 x 6.22 / 0.082 x 298
n = 0.245
Mass O₂ :
= mol x MW
= 0.245 x 32
= 7.84 g
Please help me answer this question but in your own words!!!!
Answer:
when sodium reacts with chlorine atoms, sodium will give an electron which is negative charged to chlorine.it makes sodium(+) and chlorine (-)..then sodium will attract chlorine to form sodium chloride(NaCl)
Base your answer on the information below. The hydrocarbon 2-methylpropane reacts with iodine as represented by the balanced equation below. At standard pressure, the boiling point of 2-methylpropane is lower than the boiling point of 2-iodo-2-methylpropane. Explain the difference in the boiling points of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity and intermolecular forces.
Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
what type of reaction is Au2S+H2---> 2Au+H2S
Answer:
single replacement reaction is type of reaction is Au2S+H2---> 2Au+H2S.