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Yanka [14]
3 years ago
12

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?

Chemistry
1 answer:
Anvisha [2.4K]3 years ago
3 0
In order to emit electrons, the cesium will have to absorb photons. Each photon will knock out one electron by transferring its energy to the electron. Therefore, by the principle of energy conservation, the energy of the removed electron will be equal to the energy of the incident photon. That energy is calculated using Planck's equation:

E = hf

E = 6.63 x 10⁻³⁴ * 1 x 10¹⁵
E =  6.63 x 10⁻¹⁹ Joules

The electron will have 6.63 x 10⁻¹⁹ Joules of kinetic energy
You might be interested in
How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol
netineya [11]

The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

 H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ

The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid

  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

4 0
3 years ago
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The volume of an irregular solid can be found from its displacement in a known volume of water.
Eva8 [605]

Answer:

The volume of an irregularly shaped solid can be determined from the volume of water it displaces. A graduated cylinder contains 19.9 mL of water. When a small piece of galena is added, it sinks and the volume increases to 24.5 mL

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Any method of determining whether an event or object is older or younger than others
Inga [223]
The answer is relative dating, btw

6 0
3 years ago
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Lead (11) oxide dissolves in both
inn [45]

Answer:

bacic

Explanation:

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7 0
3 years ago
How many moles of ammonium nitrate will be produced 110.0g of ammonium carbonate
marshall27 [118]

Answer:

                     2.288 Moles of NH₄NO₃

Explanation:

The Balance chemical equation is as follow:

                  (NH₄)₂CO₃ + 2 HNO₃  →  2 NH₄NO₃ + H₂O + CO₂

To solve this problem we will do following steps:

Finding moles of Ammonium Carbonate:

As we know,

                        Moles  =  Mass / M.Mass

So,

                        Moles  =  110 g  /  96.08 g/mol

                        Moles  =  1.144 moles

Calculating moles of Ammonium Nitrate:

According to balance chemical equation;

                      1 mole of (NH₄)₂CO₃ produces  =  2 moles of NH₄NO₃

So,

          1.144 moles of (NH₄)₂CO₃ will produce  =  X moles of NH₄NO₃

Solving for X,

                     X =  2 moles × 1.144 moles ÷ 1 mole

                     X  =  2.288 moles of NH₄NO₃

8 0
3 years ago
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