What does a low solubility product constant indicate about a compound?

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a

Inessa [10]

Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits is mathematically given as

N=1820slits

<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>

Generally, the equation for the wave is mathematically given as

$d\ sin\ (\theta\ m) \ = \ m\ \lambda$

Where the chromatic resolving power (R) is defined by

$R\ =\ \lambda\ / \ d \ \lambda$

R = nN,

Therefore

$\lambda_1 \ = \ (656.45)(1 \ * \ 10^{-9})/1mm$

$\lambda_1= 656.45*10^{-9}$

and

$\lambda_2= (656.27)(1*10^{-9})/1mm$

$\\\\\lambda_2= 656.27*10^{-9}m$

In conclusion, the minimum number of slits is required to resolve these two wavelengths in second-order

$N\ =\ \dfrac{\lambda}{m\ d\ T\ }\\\\$

Therefore

$N\ =\ \dfrac{656.45 \ * \ 10^{-9}}{2\ * \ (0.18*10^{-9})}$

N=1820slits

Read more about slits

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7 0

2 years ago

Write word and balanced chemical equations to show the

Alex777 [14]

Answer:

chemical laboratory Pune Maharashtra India

5 0

3 years ago

Using the periodic tables, identify the heaviest member of each of the following groups:

Lady_Fox [76]

Answer:

(a) Alkali metals: Francium (Fr)

(b) Chalcogens: Polonium (Po)

(c) Noble gases: Radon (Rn)

(d) Alkaline earth metals: Radium (Ra)

Explanation:

In the periodic table, the atomic mass increases down the group. Therefore, the last element of a group is the heaviest element of the group.

(a) alkali metals: The chemical elements that are present in group 1 of the periodic table, except hydrogen.

<u>The heaviest member of this group is francium (Fr)</u>

(b) chalcogens: The chemical elements that are present in group 16 of the periodic table

<u>The heaviest member of this group is polonium (Po)</u>

(c) noble gases: The chemical elements that are present in group 18 of the periodic table

<u>The heaviest member of this group is radon (Rn)</u>

(d) alkaline earth metals: The chemical elements that are present in group 2 of the periodic table.

<u>The heaviest member of this group is radium (Ra)</u>

7 0

4 years ago

Deval drew the models of particles in a substance shown below. Which model best represents the particles in solid

adell [148]

Answer:

it would be the second one thank me later

Explanation:

5 0

2 years ago

What is the binding energy of a nucleus that has a mass defect of 5.81*10-^29 kg

IrinaVladis [17]

Answer:

Choice A: Approximately $5.23 \times 10^{-29}$ joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the $\rm 5.81\times 10^{-29}$ kilograms of mass.

$E = m \cdot c^{2}$ ,

where

$E$ stands for energy,
$m$ stands for mass, and
$c$ is the speed of light in vacuum.

The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of $c$ ?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, $c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}$ .

Apply the mass-energy equivalence:

$\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}$ .

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

$\rm 1\; J = 1\; N \times 1\; m$ .

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

$\begin{aligned}\rm 1\; J &= \rm 1\; N \times 1\; m\\ & = \rm \left(1\; kg\times 1\; m\cdot s^{-2}\right)\times 1\; m\\ &= \rm 1\; kg \cdot m^{2}\cdot s^{-2}\end{aligned}$ .

In other words, the mass defect here is also $\rm 5.23\times 10^{-12}\; J$ .

7 0

3 years ago

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