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LenKa [72]
2 years ago
11

A photon has a frequency of 7.3 × 10–17 Hz. Planck’s constant is 6.63 × 10–34 J•s. The energy of the photon, to the nearest tent

hs place, is _____ × 10–50 J.
Chemistry
2 answers:
yawa3891 [41]2 years ago
7 0

Given:

E = 7.3 × 10–17 Hz                                                                                      

 h= 6.63 × 10–34 J•s

Now <em>E = hf</em>

where E is the energy of the photon                                                          

h is the Planck's constant                                                                          

f is the frequency of the photon

Substituting the values in the equation we get                                        

E= 7.3 × 10^-17 × 6.63 × 10^-34                                                                  

<u>E= 4.8399 × 10^-50  J. </u>                                                                                                      



Monica [59]2 years ago
6 0

Answer:

4.8

Explanation:

just took the edge exam

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A proton transfer reaction can occur when an aldehyde is placed in strong base, such as an alkoxide ion, producing an alcohol an
Pani-rosa [81]

Hi, you have not provided structure of the aldehyde and alkoxide ion.

Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.

Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.

The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.

After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.

All the structures are shown below.

7 0
2 years ago
. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture
postnew [5]

Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

3 0
2 years ago
How many grams of antifreeze C2H4(OH)2 would be required per 500 g of water to prevent the water from freezing at a temperature
Wewaii [24]

Answer:

333.7 g.

Explanation:

  • The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>

Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).

Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>

molaity (m) is the no. of moles of solute per kg of the solvent.

∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))

∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.

∵ no. of moles = mass/molar mass.

<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>

5 0
2 years ago
A volumetric pipette has an uncertainty of 0.01cm3. What are the lowest and highest possible volumes for a measurement of 0.20cm
Anarel [89]

Answer:

Possible lowest volume = 0.19 cm

Possible highest volume = 0.21 cm

Explanation:

given data

volumetric pipette uncertainty  =  0.01 cm³

total volume = 0.20 cm³

solution

we will get here Possible lowest and highest volume that is express as

Possible lowest volume = total volume - uncertainty   .....................1

Possible highest volume = total volume + uncertainty    ....................2

put here value in both equation and we get

Possible lowest volume = 0.20 cm - 0.01 cm

Possible lowest volume = 0.19 cm

and

Possible highest volume = 0.20 cm + 0.01 cm

Possible highest volume = 0.21 cm

3 0
2 years ago
The chemical reaction that causes chromium to corrode in air is given by 4Cr+3O2→2Cr2O3 in which at 298 K ΔH∘rxn = −2256 kJ ΔS∘r
MAVERICK [17]

Answer:

-2092 kJ

Explanation:

Let's consider the chemical reaction that causes chromium to corrode in air.

4 Cr + 3 O₂ → 2 Cr₂O₃

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ΔH° - T × ΔS°

where,

  • ΔH°: standard enthalpy of the reaction
  • T: absolute temperature
  • ΔS°: standard entropy of the reaction

ΔG° = -2256 kJ - 298 K × (-0.5491 kJ/K)

ΔG° = -2092 kJ

5 0
2 years ago
Read 2 more answers
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