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3 years ago

Discuss Newland octave law and it's rejection

1 answer:

7 0

It's where chemical elements are organized based on atomic weight. Also, like/similar physical and chemical properties in each interval of 7 elements.
It was a silly reason it was rejected. It was rejected because it's octaves were too similar to those in music.
I hope this helps!

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Calculate the concentration of clo2− at equilibrium if the initial concentration of hclo2 is 2. 25×10^−2 m.

jek_recluse [69]

The concentration of ClO₂⁻ at equilibrium if the initial concentration of HClO₂ is 0.0654.

<h3>What is concentration?</h3>

The concentration of any substance is the quantity of that substance in per square of the space or container.

The reaction is

HClO₂ + H₂O <=> H₃O⁺ + ClO₂⁻

The pH is 0.454 M

Ka = [H₃O⁺][ClO₂⁻ ] / [HClO₂]

2. 25 × 10⁻² m = [x][x] / 0.454-x]

2 + 0.011 - 0.004994 = 0

solve the quadratic equation

x = 0.0654 = [H3O+] = [ClO2-]

pH = -log (H3O+)

pH = -log(0.0654)

pH = 1.2

equilibrium concentrations of

[HClO2] = 0.454 -x = 0.454 -0.0654 = 0.3886 M

[ClO2- ] = x = 0.0654

Thus, the equilibrium concentrations is 0.0654.

To learn more about concentration, refer to the link:

brainly.com/question/16645766

#SPJ4

6 0

1 year ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago

Describe the process by which ag+ ions are precipitated out of solution

MA_775_DIABLO [31]

Describe the process by which Ag+ ions are precipitated out of solution. 4. In your testing, several precipitates are formed, and then dissolved as complexes.

3 0

2 years ago

Read 2 more answers

Which of the following is not the same as 0.00684 kg? 0.0684 dkg,6.84 g, 684 cg, 6,840 mg,

lesya [120]

The Answer would be C.) 684 cg

8 0

3 years ago

Read 2 more answers