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2 years ago

Which solute-solvent pair would be expected to exhibit the greatest solubility?

Chemistry

1 answer:

Maru [420]2 years ago

4 0

Answer:

butter in oil

Explanation:

Butter in oil will have the greatest solubility because they are both like compounds.

The general rule of solubility is that like will dissolve like. Similar compounds will dissolve in similar solvent.

Generally, polar solute dissolves readily in polar solvent. An example is salt and water.

Non-polar solute will dissolve readily in non-polar solvent.

Butter is a non-polar solute, oil is a non-polar solvent. Therefore, they will readily dissolve in one another.

Water is a polar solvent

Grease is non-polar solute

Table salt is polar solute

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Driving cars lowers the pH of the oceans by _______.

Anna [14]

Correct answer: a. releasing CO2 that dissolves and forms acid in the oceans

The fuels used in automobiles release gases like carbon dioxide, carbon monoxide, oxides of nitrogen and sulfur. Carbon dioxide when dissolved in water forms carbonic acid. So, when the usage of cars is high, these emissions of carbon-dioxide into the atmosphere increase and this leads to the lowering of pH of the oceans as the carbon dioxide present in higher amounts in to atmosphere diffuses into the oceanic waters and form carbonic acid which makes the ocean slightly acidic.

$CO_{2}(g) + H_{2}O (l) H_{2}CO_{3}(aq)$

6 0

2 years ago

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

2 years ago

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A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain

krek1111 [17]

Answer:

- 0.0249% Sb/cm

$-1.2465 * 10^9 \frac{atoms}{cm^3.cm}$

Explanation:

Given that:

One surface contains 1 Sb atom per 10⁸ Si atoms and the other surface contains 500 Sb atoms per 10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb per cm can be calculated as follows:

The difference in concentration = $\delta_c$

The distance $\delta_x$ = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing 1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per 10⁸ silicon atoms can be calculated as follows:

$\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%$

= - 0.0249% Sb/cm

b) The concentration $(c_1)$ of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

$c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}$

Lattice parameter = 5.4307 Å; To cm ; we have

= $5.4307A^0* \frac{10^{-8}cm}{ A^0}$

$c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}$

= $0.00499*10^{17}atoms/cm^3$

The concentration $(c_2)$ of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

$c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}$

= $\frac{4*10^{-3}}{1.601*10^{-22}}$

= $2.4938*10^{17}atoms/cm^3$

Finally, to calculate the concentration gradient

$(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}$

$(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}$

$= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}$

8 0

2 years ago

100 POINTS!!!!Questions/Results Heat energy, Q, is calculated by multiplying the mass of the object in grams, m, by the change o

zepelin [54]

Q = mΔT(Cp)
where Q = heat energy in J (joules),
m = mass in g, ΔT = change in temper. (°C),
Cp = heat capacity in J/(g°C)

Water has a higher heat capacity, meaning that once heat energy is absorbed, it holds that heat longer than bread. Also though, a higher heat capacity of water means that it takes more energy to heat it up.

I don't see any specific data listed for this lab??

5 0

2 years ago

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Consider the picture of a gas pump. Which type of gasoline has the highest percentage of octane (the main component of gasoline)

Readme [11.4K]

Answer:

premium: 91 octane rating

Explanation:

Octane number refers to the percentage or volume fraction of isooctane in a fuel.

The octane number gives a picture of how safe a fuel is for an engine. The higher the octane rating the lesser the tendency of the fuel to cause knocking of the engine.

The type of gasoline with the highest percentage of octane among the options is premium.

5 0

2 years ago

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