Is this experiment okay for this question???? If not, what would a good one be?

Calculate the wavelength of the photon emitted when an electron makes a transition from n=6 to n=3. You can make use of the foll

Angelina_Jolie [31]

Answer: The wavelength of light is $1.094\times 10^{-6}m$

Explanation:

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level = 3

$n_i$ = Initial energy level = 6

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{6^2} \right )\\\\\lambda =\frac{1}{914617m^{-1}}=1.094\times 10^{-6}m$

Hence, the wavelength of light is $1.094\times 10^{-6}m$

6 0

4 years ago

During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S

Kay [80]

Answer: The below calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of $^{235}UF_6=349.034348g/mol$

Molar mass of $^{238}UF_6=352.041206g/mol$

By taking their ratio, we get:

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}$

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}$

From the above relation, it is clear that rate of effusion of $^{235}UF_6$ is faster than $^{238}UF_6$

Difference in the rate of both the gases, $Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816$

To calculate the percentage increase in the rate, we use the equation:

$\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100$

Putting values in above equation, we get:

$\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%$

The above calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

3 0

3 years ago

Are shrubs and small animals primary secondary or both? Are small trees primary secondary or both? These are both separate quest

In-s [12.5K]

Answer:

1.

shrubs are producers.
small animals are primary if they eat plants only.
they're both if the consumer is omnivorous

2. most plants , including trees, are producers so they can't be consumers.

only rare plants like the venus fly trap is a consumer because they eat bugs.

5 0

3 years ago

Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of

never [62]

Answer:

−153.1 J / (K mol)

Explanation:

Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)

Hg(liq) + Cl2(g) → HgCl2(s)

Given that;

The standard molar entropies of the species at that temperature are:

Sºm (Hg,liq) = 76.02 J / (K mol) ;

Sºm (Cl2,g) = 223.07 J / (K mol) ;

Sºm (HgCl2,s) = 146.0 J / (K mol)

The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]

= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]

= -153.09 J / (K mol)

= or -153.1 J / (K mol)

Hence the answer is −153.1 J / (K mol)

5 0

4 years ago

Which of the following correctly describes a compound?

trasher [3.6K]

Answer: Option (a) is the correct statement.

Explanation:

A substance that has two or more number of different atoms which are chemically combined to each other in a fixed ratio by mass is known as a compound.

For example, $Ca + Cl_{2} \rightarrow CaCl_{2}$

Here, $CaCl_{2}$ contains one atom of calcium and two atoms of chlorine which are chemically combined together in 1:2 ratio. Therefore, it is a compound.

Also, both chemical and physical properties of $CaCl_{2}$ will be different from Ca and $Cl_{2}$ .

Thus, we can conclude that the statement atoms are bonded together, and the compound has different physical and chemical properties than the individual element, is correct.

8 0

3 years ago