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2 years ago

The element vanadium had a line with a wavelength of 318.5 nm in its emission spectrum. What is the frequency of this line

Chemistry

1 answer:

Hunter-Best [27]2 years ago

7 0

The frequency of this line of vanadium is 9.38 x10 ^14 Hz.

Emission spectrum shows how the electron of an atom goes or moves from a higher to a lower energy level.

Now The energy of a photon is given by

E = hc/λ

where

h = Plank's constant = 6.626 x 10⁻³⁴ J s

c = speed of light= 3 x 10⁸ m/s

λ = wavelength = 318.5 x 10⁻⁹ m

Solving

E = (6.626 x 10⁻³⁴ J s x 3 x 10⁸ m/s) / 318.5 x 10⁻⁹ m

E =6.2166 x10 ^-19 J

Also, we know that energy is related to frequency by the equation

E =hf

Where;

h = Planks's constant

f = frequency of photon

Making frequency subject of the formulae

f = E/h

f =6.2166 x10 ^-19 J/ 6.626 x 10⁻³⁴ J s

f = 9.38 x10 ^14 Hz

See similar question and solution here:brainly.com/question/24630281

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

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what is the order of decreasing bond length for a C-C bond composed of the following molecular orbitals?I. sp^3 - sp^3II. sp^2 -

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Answer:

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Explanation:

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$\text{Bond length} \propto \frac{1}{\text{Number of bonds}}$ ...[1]

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$\text{Bond Energy} \propto \text{Number of bonds}$ ..[2]