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3 days ago

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If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series em

ission spectrum. Calculate the wavelength λ associated with the fifth line

1 answer:

castortr0y [2.7K]3 days ago

7 0

Response: The wavelength tied to the fifth line is 397 nm

Clarification:

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of emitted radiation

E= energy

For the fifth line in the hydrogen atom's spectrum within the Balmer series, the transition occurs from n=2 to n=7.

Applying Rydberg's Equation:

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

Where,

$\lambda$ = Wavelength of radiation

Rydberg's Constant = $R_H$ $10973731.6m^{-1}$

$n_f$ = Higher quantum level = 7

$n_i$ = Lower quantum level = 2 (within Balmer series)

Substituting the values into the equation above, we obtain

$\frac{1}{\lambda}=10973731.6\times \left(\frac{1}{2^2}-\frac{1}{7^2} \right )\times 1^2$

$\frac{1}{\lambda}=2.52\times 10^{6}m$

$\lambda}=3.97\times 10^{-7}m=397 nm$ $1nm=10^{-9}m$

Therefore, the wavelength λ corresponding to the fifth line is 397 nm

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HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

Alekssandra [2711]

Answer:

Quantity of $H_{2}O$ generated will be reduced to fifty percent of its initial amount.

Explanation:

Equilibrium reaction: $HCl+NaOH\rightarrow NaCl+H_{2}O$

In accordance with the balanced equation, 1 mol of HCl interacts with 1 mol of NaOH leading to the formation of 1 mol of $H_{2}O$

<pif the="" quantities="" of="" reactants="" and="" hcl="" are="" diminished="" by="" half="" it="" results="" in:="">

0.5 mol of HCl interacting with 0.5 mol of NaOH yielding 0.5 mol of $H_{2}O$ .

Consequently, it is clear that the total of $H_{2}O$ produced will be halved if the quantities of the reactants are halved.

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5 0

1 month ago

Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a

lions [2653]

Answer:

Explanation:

The relationship between the new temperature scale and the absolute temperature scale is defined as follows

Aw = 2 K

for K = 273.15 (the freezing point of water on the absolute scale)

Aw = 2 x 273.15 = 546.3 K

Each division of the new scale is equivalent to half that of each division on the absolute scale

each division of the new scale is minimal.

The value of R = 8.314 J per mole per K

Here, per K corresponds to 2Aw

Hence, the value of R in the new scale = 8.314/2 J per mole per Aw

= 4.157 J per mole per Aw

k = R / N

= 4.157 / 6.02 x 10²³

= .69 x 10⁻²³

= 6.9 x 10⁻²⁴ J per molecule per Aw .

7 0

1 month ago

Read 2 more answers

The speed of light in a vacuum is 2.998 × 10 8 8 m/s. How long does it take for light to circumnavigate the Earth, which has a c

castortr0y [2743]

Answer:- 0.134 seconds

Solution:- The speed is given as $2.988*10^8\frac{m}{s}$ and the circumference is 24900 miles which is same as the distance light have to covered. It asks to calculate the time required to cover this distance by the light.

Unit conversion is needed from miles to meters since the speed is given in meters per second.

1 mile = 1609.34 meters

Thus, $24900mile(\frac{1609.34meter}{mile})$

= 40072566 meters

Now, $speed=\frac{distance}{time}$

Rearranged for time, that gives: $time=\frac{distance}{speed}$

Inserting the values:

$time=\frac{40072566meter}{2.988*10^8\frac{meter}{second}}$

= 0.134 seconds

Hence, light would take 0.134 seconds to traverse the indicated distance. The answer without the unit is 0.134.

8 0

1 month ago

En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

KiRa [2711]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En esta cuestión analizaremos el millón de litros de agua anualmente, dado que este dato nos permite calcular el calor necesario para calentar dicha cantidad, considerando que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

A continuación, utilizamos la entalpía de combustión del metano para determinar la cantidad en kilogramos necesaria, ya que la energía calórica perdida por el metano es equivalente a la energía obtenida por el agua:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este supuesto, tenemos que, bajo condiciones normales de 1 bar y 273 K, el precio de 1 metro cúbico de metano es 0,45 €, lo que nos permite calcular las moles de metano en esas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

En consecuencia, los kilogramos de metano que se obtienen por 0,45 € son:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Finalmente, usando regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

29 days ago

How many atoms of Mg are present in 97.22 grams of Mg?

VMariaS [2690]

Answer: The correct choice is (b).

Explanation:

We have the mass of Magnesium at 97.22 g, and the molar mass of Magnesium is 24.305 g/mol.

Therefore, the calculation for the number of moles is as follows.

Number of moles = $\frac{mass given in grams}{Molar mass}$

= $\frac{97.22 g}{24.305 g/mol}$

= 4 mol

Additionally, it is known that one mole contains $6.023 \times 10^{23}$ atoms/mol. Thus, we calculate the total number of atoms in 4 moles as follows.

$4 mol \times 6.023 \times 10^{23} atoms/mol$

= $24.08 \times 10^{23}$ atoms

or, = $2.408 \times 10^{23}$ atoms

Hence, we conclude that in 97.22 grams of Magnesium, there are $2.408 \times 10^{23}$ atoms.

7 0

25 days ago