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2 years ago

What is the frequency of radiation whose wavelength is 2.51 x 10 -7m?

Chemistry

1 answer:

hodyreva [135]2 years ago

8 0

The frequency of radiation : 1.2 x 10¹⁵ Hz

<h3>Further explanation </h3>

Radiation energy is absorbed by photons

The energy in one photon can be formulated as

$\large{\boxed{\bold{E\:=\:h\:.\:f}}}$

Where

h = Planck's constant (6,626.10⁻³⁴ Js)

f = Frequency of electromagnetic waves

c = speed of light

= 3.10⁸ m/s

λ = wavelength

Wavelength-λ is 2.51 x 10⁻⁷m

The frequency :

$\tt f=\dfrac{3.10^8}{2.51\times 10^{-7}}=1.2\times 10^{15}$

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Assume that the pressure inside of the 1 liter balloon is 200,000 Pa, what is the explosive energy in the inside the balloon? Ex

AysviL [449]

Answer:

200 Joules is the explosive energy in the inside the balloon. And that is $9.523\times 10^{-5}$ 1 lb of TNT.

Explanation:

$Pressure=\frac{force}{Area}=\frac{Force\times distance}{Area \times distance}=\frac{Energy}{Volume}$

Volume of the balloon = V = 1 L = $0.001 m^3$

Pressure inside the balloon ,P= 200,000 Pa = $200,000 N/m^2$

Explosive energy in the inside the balloon be E.

E = Pressure × Volume

$E=200,000 N/m^2\times 0.001 m^3=200 Joules$

1 lb of TNT = $2.1\times 10^6 J$

200 Joules = $200\times \frac{1}{2.1\times 10^6 }$ 1 lb of TNT

= $9.523\times 10^{-5}$ 1 lb of TNT

5 0

2 years ago

10.

Andru [333]

Answer: 1.8 moles Fe and 2.7 moles $CO_2$ are produced.

Explanation:

The balanced chemical reaction is:

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

According to stoichiometry :

3 moles of $CO$ require 1 mole of $Fe_2O_3$

Thus 2.7 moles of $CO$ will require= $\frac{1}{3}\times 2.7=0.9moles$ of $Fe_2O_3$

Thus $CO$ is the limiting reagent as it limits the formation of product and $Fe_2O_3$ is the excess reagent.

As 3 moles of $CO$ give = 2 moles of $Fe$ and 3 moles of $CO_2$

Thus 2.7 moles of $CO$ will give = $\frac{2}{3}\times 2.7=1.8moles$ of $Fe$ and $\frac{3}{3}\times 2.7=2.7moles$ of $CO_2$

Thus 1.8 moles Fe and 2.7 moles $CO_2$ are produced.

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2 years ago

HELP! THIS IS A TIMED QUIZ!!

ivanzaharov [21]

Answer:

A) Age!! Its because even if the object was 1 year old or 100 years old, nothing about the impact would change. However, those other categories depict features that would definitely make an impact. For example as object that is big, fast, and hits at an angle perpendicular to whatever it is moving towards, the impact will be very lage. But if its the opposite and it was small and slow, then the impact crater would not be as large. Good luck on your quiz!!

5 0

2 years ago

A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt

never [62]

Answer:

a) Δπ = 1.264 atm

b) W = 128 joules

c) ΔH >> W ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

R= 0.0821 Latm/kmol

T = 300 K

calling π(sw) osmotic pressure for for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W = 101.33 j/ Latm x 1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

Q₁ = heat required to bring the solution from 300 K to boiling, 373 K

ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH = 0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0

3 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

2 years ago

What is the frequency of radiation whose wavelength is 2.51 x 10 -7m?​

What is the frequency of radiation whose wavelength is 2.51 x 10 -7m?