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2 years ago

10

A student drops some metal paper clips in a box of sand. The student uses a magnet to separate the metal paper clips from the sa

nd. The reason this works is that the sand and paper clips together are a
A) mixture.
B) element.
C) compound.
D) substance.

1 answer:

Dmitriy789 [7]2 years ago

3 0

Answer:

compound

Explanation:

because now there are one together

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2 years ago

Rank the following in terms of decreasing miscibility in C₈H₁₈ (octane), a major component of gasoline:

Aleksandr-060686 [28]

Answer:

In order of decreasing miscibility

C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)

Explanation:

The solubility of a solid is a measure of its ability to dissolve in a liquid while for liquids, the miscibility is a measure of thhe liquid to mix with anoyjer liquid resulting in a soltion which can hold any amount of either liquids. Immiscible liquids are those that are not soluble or have very limited solibility with each other.

C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)

In the order of decreasing miscibility as like dissolve like, ability to dissociate and polar and organic characteristics are considered

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2 years ago

What would happen to the water’s boiling point if MgSO4 is added to it?

kirill [66]

Answer:

When Epsom salt (magnesium sulfate) dissolves, it separates into its ions: a magnesium ion (Mg2+) and a sulfate ion (SO4 2-), which results in hard water. When hard water and soap are mixed, the magnesium ion reacts with soap molecules and forms a solid material called a precipitate, which does not dissolve.

Explanation:

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2 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

2 years ago

18. What is the percent by mass sugar of a solution made by dissolving 12.45 grams of sugar in 100 grams of water?

Fantom [35]

The percent by mass sugar of a solution : 11.07%

<h3>Further explanation</h3>

Given

mass of sugar = 12.45 g

mass of water = 100 g

Required

The percent by mass

Solution

Mass of solution :

$\tt mass~sugar+mass~water=12.45+100=112.45~g$

Percent mass of sugar :

$\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%$

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2 years ago