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agasfer [191]
3 years ago
8

Microwave ovens work by irradiating food with microwaves, which are absorbed by the water molecules in the food and converted to

heat. Assuming that microwave ovens emit radiation with a wavelength of 14.8 cm, calculate how many photons are emitted per second in a 1 x 103-Watt microwave oven? (1 W = 1 J/s)
Chemistry
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

7.45\times 10^{26} photons are emitted per second in a 1\times 10^3 Watt microwave oven.

Explanation:

The expression for the power is:-

Power=\frac{Energy}{Time}

Also, E=n\times \frac{h\times c}{\lambda}

Where,  

n = the number of photons

h =  Plank's constant having value 6.626\times 10^{-34}\ Js

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

\lambda = the wavelength of the light

So,  

Power=\frac{n}{Time}\times \frac{h\times c}{\lambda}

Thus, the expression for photons per second is:-

\frac{n}{Time}=\frac{Power\times \lambda}{h\times c}

Given that:-

Power = 10³ Watt

( 1 cm = 0.01 m)

[/tex]

So,  

Photon/sec=\frac{10^{3}\times 0.148}{6.626\times 10^{-34}\times 3\times 10^8}=7.45\times 10^{26}

7.45\times 10^{26} photons are emitted per second in a 1\times 10^3 Watt microwave oven.

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The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r
Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0
3 years ago
In a reaction between 6.0 g of oxygen gas, 4.0 g of hydrogen gas, and 5.0 g of solid sulfur at standard temperature and pressure
nikitadnepr [17]

Answer:

The limiting reagent is the O₂

Explanation:

We can think, this reaction

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6 g / 32 g/m = 0.187 mole O₂

4g / 2 g/m = 2 mole H₂

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Ratio between reactants is 2:1:1, 1:2:1, 1:1:2

For 2 mole of O₂, I need to react 1 mol of H₂ and 1 mol of S

0.187 mole of O₂, I need (the half)

0.093 mole of H₂ and 0.093 mole of S

For 1 mole of H₂, I need to react 2 mole of O₂ and 1 mol of S

2 mole of H₂, I need (the double of O₂ and the same for S)

4 mole of O₂ ; 2 mole of S

For 1 mol of S, I need to react 1 mol of H₂ and 2 mole of O₂

0.156 mole I need the same amount for H₂ and the double for O₂

0.156 mole of H₂ and 0.312 mole of O₂

In both cases, I can't make react, all the mass of oxygen, so this is the limiting reagent.

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