Answer:
Mg S2 O3
Explanation:
.691 g of Mg is .284 mole
1.84 g of S is .5739 mole
1.365 g of O is .8531 mole you can see the ratio is ~ 1 :2 :3
Mg S2 O3
Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are known as colligative properties. For this problem, we use boiling point elevation concept.
ΔT(boiling point) = (Kb)mi
ΔT(boiling point) = (0.51 C-kg / mol )(4.0 mol / 2.05 kg ) (2)
ΔT(boiling point) = 1.99 C
T(boiling point) = 101.99 C
Answer:
C. porous
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Answer:
There are two types of hydrocarbons: aliphatic and aromatic. The three types of aliphatic hydrocarbons are alkanes, alkenes, and alkynes. Aromatic hydrocarbons include benzene. Overall, examples of hydrocarbons are methane, ethane, propane, and butane.
