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3 years ago

Anhydrous calcium chloride is used after washing your organic product with water. the purpose of this step is to:

2 answers:

7 0

The purpose of washing the organic product that contains water with anhydrous calcium chloride is to remove water from the organic product, thereby, increasing its purity. Calcium chloride is known to be one of the most effective drying agents together with sodium sulfate and magnesium sulfate. It is especially used for hydrocarbons. It is an inorganic salt that readily takes up the water from the organic product.

Brrunno [24]3 years ago

5 0

Anhydrous solids are the crystallized crystals when the hydrate form of the compound undergoes crystallization. Hence, when an anhydrous salt is placed in an aqueous environment, it will absorb water in order to be a hydrate once again. Thus, anhydrous calcium chloride is used to absorb the water to separate it from the organic product.

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The molar mass of SO3 is grams. in significant figures

Vikentia [17]

Answer:

80.066 g/mol

Explanation:

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3 years ago

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Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as

Sonbull [250]

Answer:

0.100 m AlCl3 will have the highest boiling point

Explanation:

Step 1: Data given

The molal boiling point elevation constant for water is 0.51°C/m

Since those are all aqueous solutions, the have the same molal boiling point elevation constant

Step 2:

0.100 m C6H12O6

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr = 1

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT = 1 * 0.51 * 0.100

ΔT = 0.051 °C

0.100 m NaCl

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr = Na+ + Cl- = 2

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =2 * 0.51 * 0.100

ΔT = 0.102 °C

0.100 m AlCl3

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr = Al^3+ + 3Cl- = 4

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =4 * 0.51 * 0.100

ΔT = 0.204 °C

0.100 m MgCl2

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr = Mg^2+ +2Cl- = 3

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =3 * 0.51 * 0.100

ΔT = 0.153 °C

0.100 m AlCl3 will have the highest boiling point

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3 years ago

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disa [49]

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2 years ago

A solution of water (kf=1.86 ∘c/m) and glucose freezes at − 2.15 ∘c. what is the molal concentration of glucose in this solution

Reil [10]

<span>1.16 moles/liter The equation for freezing point depression in an ideal solution is Î”TF = KF * b * i where Î”TF = depression in freezing point, defined as TF (pure) â’ TF (solution). So in this case Î”TF = 2.15 KF = cryoscopic constant of the solvent (given as 1.86 âc/m) b = molality of solute i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1. Solving for b, we get Î”TF = KF * b * i Î”TF/KF = b * i Î”TF/(KF*i) = b And substuting known values. Î”TF/(KF*i) = b 2.15âc/(1.86âc/m * 1) = b 2.15/(1.86 1/m) = b 1.155913978 m = b So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>

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3 years ago

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RUDIKE [14]

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3 years ago

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