How does the rate of temperature change depend on the temperature difference of the two beakers?

Nitrous acid, hno2, has a ka of 7. 1 ✕ 10-4. what are [h3o ], [no2-], and [oh -] in 0. 40 m hno2?

romanna [79]

Nitrous acid, hno2, has an acid dissociation constant - ka of 7. 1 ✕ 10-4. what are [h3o ], [no2-], and [oh -] in 0. 40 m hno2 - 4829 M [OH^-] = 1.439 x 10^-14 M

The acid dissociation constant (Ka) is used to differentiate between strong and weak acids. Strong acids have very high Ka values. The Ka value is determined by examining the equilibrium constant for acid dissociation. The acid dissociates more readily as the Ka increases.

The original molecular definition of an acid, according to Arrhenius, is a molecule that dissociates in an aqueous solution, releasing the hydrogen ion H+ (a proton): HA A + H+. acid dissociation constant is an equilibrium constant for this dissociation reaction.

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4 0

1 year ago

I will give brainlyest<br> Explain why science is a continuous progression of study.

Likurg_2 [28]

It’s because new discoveries are made all the time sometimes what was considered right may be found out to be wrong

8 0

3 years ago

Describe the photoelectric effect and explain why it made modifications to the Rutherford model necessary.

Furkat [3]

<span>Photoelectric effect refer to the emission of electron or free carriers when light shine into a material. According to the Rutherford model, light of any energy should be able to make electrons leave the atom and be emitted. The energy of the emitted electrons should be related to the intensity of the light. But the energy of the electron is actually only related to the energy of the light</span>

6 0

3 years ago

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Solve the quadratic equation 2x^2+13x=15 by method of completing the square<br>

mafiozo [28]

Answer:

x = 1, -7.5

Explanation:

2x² + 13x = 15

Divide the left side of the equation by 2

2(x² + 6.5x) = 15

Divide 6.5 by 2 and square the quotient

6.5/2 = 3.25

3.25² = 10.5625

Add 10.5625 to the left side

2(x² + 6.5x + 10.5625) = 15

Since you have a 2 outside the parentheses, you will be adding 10.5625•2 to the right side.

10.5625 • 2 = 21.125

2(x² + 6.5x + 10.5625) = 36.125

To factor (x² + 6.5x + 10.5625), add b/2 to x

b/2 = 6.5/2 = 3.25

2(x + 3.25)² = 36.125

Divide by 2

(x + 3.25)² = 18.0625

Square root.

(x + 3.25) = √18.0625

x + 3.25 = ±4.25

Subtract 3.25.

x = 4.25 - 3.25 = 1

x = -4.25 - 3.25 = -7.5

x = 1, -7.5

4 0

3 years ago

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

3 years ago

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