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pashok25 [27]
2 years ago
9

Calculate thr number of photons having a wavelength of 10.0 μm required to produce 1.0 kJ of energy.

Chemistry
2 answers:
slega [8]2 years ago
7 0

Answer:

The number of photons are 5.028\times 10^{27}.

Explanation:

E=\frac{h\times c}{\lambda}

where,

E = energy of photon =  

h = Planck's constant = 6.63\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 10.0 μm  =10^{-5} m

1 μm = 10^{-6} m

E=\frac{6.63\times 10^{-34}Js\times 3\times 10^8m/s}{10^{-5} m}

E=1.989\times 10^{-20} Joules

Let the n number of photons with energy equal to E' = 1.0 kJ = 1000 J

n\times E=E'

n\times 1.989\times 10^{-20} J=1000 J

n=\frac{1000 J}{1.989\times 10^{-20} J}=5.028\times 10^{27}

The number of photons are 5.028\times 10^{27}.

ddd [48]2 years ago
4 0
Ok so first you need to figure out the energy of ONE photon with that wavelength. Using E=hc/lambda, you get E= 1.99 * 10^-20 J/photon. Now, how many photons do you need to add up to get to one kilojoule=1000 joules? 1000J / (1.99 * 10^-20 J/photon) = approximately 5 * 10^22 photons hope this helps
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A 7.591-9 gaseous mixture contains methane (CH4) and butane
mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y <em>(1)</em>

<em>Where X is the mass of methane and Y the mass of butane</em>

<em />

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

<em>2 moles of oxygen react per mole of methane</em>

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

<em>13/2 moles of oxygen react per mole of methane</em>

<em />

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y <em>(2)</em>

<em>Where 16.04 and 58.12 are molar masses of methane and butane</em>

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

<h3>65.71%</h3>

7 0
2 years ago
50 kg of N2 gas and 10kg of H2 gas are mixed to produce NH3 gas calculate the NH3gas formed. Identify the limiting reagent in th
statuscvo [17]

Answer:

1. H2 is the limiting reactant.

2. 56666.67g ( i.e 56.67kg) of NH3 is produced.

Explanation:

Step 1:

The equation for the reaction. This is given below:

N2 + H2 —> NH3

Step 2:

Balancing the equation.

N2 + H2 —> NH3

The above equation can be balanced as follow :

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

N2 + H2 —> 2NH3

There are 6 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 3 in front of H2 as shown below

N2 + 3H2 —> 2NH3

Now the equation is balanced.

Step 3:

Determination of the masses of N2 and H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

N2 + 3H2 —> 2NH3

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

From the balanced equation above,

28g of N2 reacted with 6g of H2 to produce 34g of NH3

Step 4:

Determination of the limiting reactant. This is illustrated below:

N2 + 3H2 —> 2NH3

Let us consider using all the 10kg (i.e 10000g) of H2 to see if there will be any left of for N2.

From the balanced equation above,

28g of N2 reacted with 6g of H2.

Therefore, Xg of N2 will react with 10000g of H2 i.e

Xg of N2 = (28 x 10000)/6

Xg of N2 = 46666.67g

We can see from the calculations above that there are leftover for N2 as only 46666.67g reacted out of 50kg ( i.e 50000g) that was given. Therefore, H2 is the limiting reactant.

Step 5:

Determination of the mass of NH3 produced during the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 10000g of H2 will react to produce = ( 10000 x 34)/6 = 6g of 56666.67g of NH3.

Therefore, 56666.67g ( i.e 56.67kg) of NH3 is produced.

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2 years ago
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ziro4ka [17]

Answer:

The reason for the suspicion was because the manner in which iodine reacted chemically as well as its other chemical properties, indicated that it belonged in the same group as chlorine and bromine, while the much heavier tellurium should be placed in the previous group

The suspicion was proved to be correct when the atomic number of tellurium was found to be 52 and that of iodine was found to be 53 by later scientists

Explanation:

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2 years ago
In a chemical reaction that takes place at a fixed pressure and volume, the enthalpy change (ΔH) is –585 kJ/mol. Will this react
Molodets [167]

by sign convention ΔH is negative it means an exothermic reaction where the heat is lose so the temperature decreases.

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2 years ago
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2 physical changes that happen in aquaponics
maks197457 [2]

Answer:

The physical and chemical change that occurs in the aquaponics are given below.

Explanation:

The plants and animals grow in size and decrease the mass of plant due to eating by the fishes is a physical changes which occurs in aquaponics. The sunlight has a heat energy which is absorb by the plants present in aquaponics which is a type of endothermic reaction. In aquaponics, the ammonia present in water is converted into nitrates which is used by the plants as a nutrients. When the mass is converted into energy, it increases the temperature of the ecosystem and also the earth surface. For example, if a wood is burn, it change into heat energy which increases the temperature and cause the global warming on the earth surface.

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