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3 years ago

List two ways in which heterogeneous mixtures differ from elements.

1 answer:

3 0

Heterogeneous mixtures differ from elements in two ways. First, elements are not a mixture at all; they are pure substances. Heterogeneous mixtures are not pure. The other way is that a heterogeneous mixture can be separated by physical means; elements can't be separated at all; if the elements are separated, it will change their composition. Hope this helps

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Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.

Rudiy27

For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol

m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg

For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius

5 0

3 years ago

A plane mirror is useful for seeing:

AURORKA [14]

C. A nearly identical image.

6 0

3 years ago

Why potasium more reactive than lithium and sodium

frez [133]

So potassium is more reactive than lithium because the outer electron of a potassium atom is further from its nucleus than the outer electron of a lithium atom. Hope this answers the question. Have a nice day. Feel free to ask more questions.

5 0

3 years ago

PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC

Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Initially 0.24 M 1.50M 0.12 M

React x x x

Some amount of compound has reacted during the process.

In equilibrium we have

0.24 - x 1.50 - x 0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

5 0

3 years ago

Research benzene and list two modern-day commercial uses for his chemical

Juli2301 [7.4K]

Benzene is used for pesticides and detergents. It is also used for other things besides these.

8 0

3 years ago