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Nitella [24]
3 years ago
6

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]3 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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ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
3 years ago
An ideal gas sample is confined to 3.0 L and kept at 27 °C. If the temperature is raised to 77 °C and the initial pressure was 1
Nadusha1986 [10]

The gas is confined in 3.0 L container ( rigid container) ⇒ the volume remains constant when the temperature is increased from from 27oC to 77oC and therefore V1=V2 .

<h2>Hope it helps you please mark as brainlist</h2>

6 0
3 years ago
Lons can have a positive or negative charge.<br> True or false
fiasKO [112]
Yes, it is true that ions can have a positive or negative charge.
7 0
4 years ago
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Calculate the amount of heat needed to boil 64.7 g of benzene (C6H6), beginning from a temperature of 41.9 C . Round your answer
JulsSmile [24]

Answer: The amount of heat needed is = 4.3kJ

Explanation:

Amount of heat H = M × C × ΔT

M= mass of benzene = 64.7g

C= specific heat capacity = 1.74J/gK

ΔT = T2-T1

Where T1 is initai temperature = 41.9C

T2 is the final temperature( boiling point of benzene) = 80.1C

H= 64.7×1.74×80.7

H= 4300J

H=4.3kJ

Therefore, the amount of heat needed is 4.3kJ

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Answer:

Option B: It remains unused during the reaction

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