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2 years ago

Calculate the number of photons having a wavelength of 12.0 μm required to produce 1.0 kJ of energy.

Chemistry

1 answer:

nika2105 [10]2 years ago

4 0

Answer:

N = 603.32 × 10 ²⁰

Explanation:

Given data:

Wavelength of photon = 12.0 μm or 12 × 10 ⁻⁶ m

Energy = 1.0 KJ or 1000 j

Number of photons = ?

Solution:

Formula:

E = n . h . f

E = energy

N = number of photons

h = planck's constant

f = frequency

it is known that

f = c/λ

thus,

E = N . h . c / λ

Now we will put the values in formula.

1000 j = ( N. 6.63 × 10 ⁻³⁴ j.s × 3 × 10⁸ m/s ) / 12 × 10 ⁻⁶ m

N = 1000 j ×12 × 10 ⁻⁶ m / 6.63 × 10 ⁻³⁴ j.s × 3 × 10⁸ m/s

N = 12000 j × 10 ⁻⁶ m / 19.89 × 10 ⁻²⁶ j. m

N = 603.32 × 10 ²⁰

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Which statement describes a change that occurs during a chemical reaction? A. Atoms in the original substances are changed into

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Answer:

B. Atoms in the original substances are arranged in a different way to make new substances.

Explanation:

The best statement that describes a change that occurs in a chemical reaction is that atoms in the original substances are arranged in a different way to make new substances.

Chemical reactions obey the law of conservation of matter.
The law postulates that "matter is neither created nor destroyed in a chemical reaction but they are simply rearranged".
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2 years ago

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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

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Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

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3 years ago

How many moles are in 4g of hydrogen?

Afina-wow [57]

Answer:

2 mole

Explanation:

Hydrogen has 2 atoms per molecule

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2 years ago

True or False - when setting up your conversion problems every numerical value must have a unit on it.

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2 years ago

Problem Page Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 70. g of

shusha [124]

Answer:

The maximum mass of carbon dioxide that could be produced by the chemical reaction is 70.6gCO_{2}

Explanation:

1. Write down the balanced chemical reaction:

$2C_{6}H_{14}_{(l)}+19O_{2}_{(g)}=12CO_{2}_{(g)}+14H_{2}O_{(g)}$

2. Find the limiting reagent:

- First calculate the number of moles of hexane and oxygen with the mass given by the problem.

For the hexane:

$70.0gC_{6}H_{14}*\frac{1molC_{6}H_{14}}{86.2gC_{6}H_{14}}=0.81molesC_{6}H_{14}$

For the oxygen:

$81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}$

- Then divide the number of moles between the stoichiometric coefficient:

For the hexane:

$\frac{0.81}{2}=0.41$

For the oxygen:

$\frac{2.54}{19}=0.13$

- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.

3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:

The calculations must be done with the limiting reagent, that is the oxygen.

$81.3gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{12molesCO_{2}}{19molesO_{2}}*\frac{44.0gCO_{2}}{1molCO_{2}}=70.6gCO_{2}$

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3 years ago