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10 days ago

The equation for the pH of a substance is pH = –log[H+], where H+ is the concentration of hydrogen ions. A basic solution has a

pH of 11.2. An acidic solution has a pH of 2.4. Based from the given, what is the approximate difference in the concentration of hydrogen ions between the two solutions? A. 1.6*10^-9 B. 4.0*10^-3 C. 6.7*10^-1 D. 1.6*10^-11

Chemistry

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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

KiRa [2933]

To determine the packing factor, begin by calculating the area and volume of the unit cell.

The area is found using:

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

In this case, R represents the radius and is connected to a as shown:

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

Substituting the value into the area formula,[ [TAG_14]]

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

The value of a is 0.4961 nm

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

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

Substituting the value,[ [TAG_27]]

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

Next, the volume can be calculated by the following method:

$V=Area\times c$

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

Inserting the value,[ [TAG_40]]

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

Now, to determine the number of atoms in the unit cell, the following equation can be employed:

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

In this context, A represents the atomic mass of $Cr_{2}O_{3}$ which is 151.99 g/mol.

Inserting all the necessary values,[ [TAG_53]]

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

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

Given, there are 2 chromium atoms and 3 oxygen atoms, thus, the total units for chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

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

Transforming them into centimeters:

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

Therefore,

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

and,

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

The total volume of the sphere will be the combined volume of all cations and anions, thus,

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

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

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

Inputting the respective values,[ [TAG_93]]

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

The atomic packing factor reflects the ratio of the volume of spheres to the volume of the crystal, so,[ [TAG_98]]

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

Thus, the atomic packing factor equals 0.758.

6 0

1 month ago

Read 2 more answers

A generic element, Z, has two isotopes, 45Z and 47Z, and an average atomic mass of 45.36 amu. The natural abundances of the two

lorasvet [2795]

The isotopic mass of 47Z is calculated to be 46.96 amu. Isotopes of a single element differ in neutron count, and to ascertain the relative atomic mass, we consider each isotope's mass weighted by their natural abundance. This provided a computation to derive the mass of 47Z.

5 0

1 month ago

How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?

Anarel [2989]

The density of human fat is $0.918 g/cm^{3}$ . The mass of the pure fat is 11.1 lbs.

First, convert the mass from pounds to grams as follows:

1 lb=453.592 g

Thus,

$11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g$

Density is defined as mass per unit volume, meaning volume can be calculated as:

$V=\frac{m}{d}$

By substituting the values,

$V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}$

Consequently, the volume gained by the individual will be $5484.61 cm^{3}$ .

6 0

2 months ago

Cathodic protection of a metal pipe against corrosion usually entails __________.

VMariaS [2998]

Answer:

C. connecting an active metal to designate the pipe as the cathode in an electrochemical cell.

Explanation:

Cathodic protection involves a method to manage the accelerated corrosion of a metal surface by designating it as the cathode within an electrochemical cell. This is accomplished by attaching the protected metal to a more sacrificial metal, which acts as the anode.

This method helps to preserve the metal by introducing a highly reactive metal that serves as the anode, supplying free electrons. By adding these free electrons, the active metal gives up its ions, protecting the less reactive steel from corrosion.

3 0

2 months ago

The ideal gas law tends to become inaccurate when Group of answer choices the pressure is raised and the temperature is lowered.

VMariaS [2998]

Answer: Option (a) is the correct answer.

Explanation:

Under conditions of low pressure and high temperature, gas molecules exhibit negligible attractions or repulsions among themselves. Hence, gases behave ideally in these scenarios.

Conversely, at low temperatures, there is a reduction in the kinetic energy of gas molecules, while high pressure compels the molecules to be closer together.

Thus, attractive forces emerge between molecules in conditions of low temperature and high pressure, causing gases to be termed real gases.

Therefore, we conclude that the ideal gas law becomes less accurate when pressure increases and temperature decreases.

5 0

2 months ago