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yanalaym [24]
4 years ago
6

Green plants absorb sunlight to power photosynthesis, the chemical synthesis of food from water and carbon dioxide. The compound

responsible for light absorption and the color of plants, chlorophyll, strongly absorbs light with a wavelength of 430 nm. Calculate the frequency of this light. Round your answer to 3 significant digits.
Chemistry
1 answer:
tekilochka [14]4 years ago
4 0

Answer: The frequency of this light is 6.74\times 10^{17}Hz.

Explanation:-

To calculate the frequency of light, we use the equation:

\lambda=\frac{c}{\nu}

where,

\lambda = wavelength of the light  =430nm=430\times 10^{-9}m      1nm=10^{-9}m

c = speed of light = 2.9\times 10^8m/s

\nu = frequency of light = ?

\nu=\frac{c}{\lambda}

\nu=\frac{2.9\times 10^8m/s}{430\times 10^{-9}m}

\nu=\frac{2.9\times 10^8m/s}{430\times 10^{-9}m}=6.74\times 10^{17}s^{-1}=6.74\times 10^{17}Hz

The frequency of this light is 6.74\times 10^{17}Hz.

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Are bf3 and nf3 symmetrical or unsymmetrical molecules? explain.
Degger [83]

The correct option is this: BF3 IS SYMMETRICAL WHILE NF3 IS UNSYMMETRICAL.

Symmetry is used in chemistry to explain the structure of chemical substances, that is, it is used to explain how the particles in chemical substances are arranged. In chemistry, molecular symmetry is used to explain the behavior of chemical substances. A chemical substance is said to be symmetrical if its polar bonds are evenly distributed; it is described as unsymmetrical if its bonds are unequally distributed. In the compound BF3, the bond dipoles cancel one another out because the polar bonds are evenly distributed whereas in NF3, the bond dipoles were not able to cancel each other out because of unequal distribution of the polar bonds, the compound has a net dipole moment.

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4 years ago
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At which temperature will the least amount of time pass before someone detects the odor of a fragrant flower?
NARA [144]
0C, 8C, 29C, 15C. I think 
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4 years ago
Two aqueous sulfuric acid solutions containing 20.0 wt% H2SO4
kozerog [31]

Answer:

The feed ratio (liters 20%  solution/liter 60% solution) is 3,08

Explanation:

In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.

100 kg of 20% solution are 100kg/1,139 kg/L = <em>87,8 L </em>

100kg×20wt% = 20 kg H₂SO₄. In moles:

20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ <em>203,9 mol</em>

The final molarity 4,00M comes from:

\frac{203,9 moles+ Xmoles}{87,8L + Yliters} <em>(1)</em>

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L×\frac{1,498 kgsolution}{L}×\frac{60 kg H_{2}SO_{4}}{100kgSolution}×\frac{1kmol}{98,08kg H_{2}SO_{4}} = <em>0,9164 kmolH₂SO₄ ≡ 916,4 moles</em>

That means:

X/Y = 916,4/100 = 9,164 <em>(2)</em>

Replacing (2) in (1):

Y(liters of 60,0 wt% H₂SO₄) = <em>28,52 L</em>

Thus, feed ratio (liters 20%  solution/liter 60% solution):

87,8L/28,52L = <em>3,08</em>

I hope it helps!

8 0
3 years ago
When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample
andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

\Delta T = change in temperature = 1.643^oC

Now put all the given value in the above formula, we get:

26380J/g=0.235g\times c\times 1.643^oC

c=68323.38J/^oC

Thus, the specific heat of calorimeter is 68323.38J/^oC

Now we have to calculate the heat of combustion of caffeine.

Formula used :

Q=c\times \Delta T

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = 68323.38J/^oC

\Delta T = change in temperature = 1.584^oC

Now put all the given value in the above formula, we get:

Q=68323.38J/^oC\times 1.584^oC

Q=108224.23J=108.2kJ

Now we have to calculate the moles of caffeine.

\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

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White raven [17]
The answer would be potassium nitrate.
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