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liubo4ka [24]
3 years ago
12

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

d responsible for light absorption and the color of plants, chlorophyll, strongly absorbs light with a wavelength of 453.nm. Calculate the frequency of this light.
Chemistry
1 answer:
Serggg [28]3 years ago
3 0

Answer: The frequency of this light is 6.62\times 10^{14}s^{-1}

Explanation:

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

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

where,

\lambda = wavelength of the light =453nm=453\times 10^{-9}m

c = speed of light = 3.0\times 10^8m/s

\nu = frequency of light = ?

\nu=\frac{3.0\times 10^8m/s}{453\times 10^{-9}m}=6.62\times 10^{14}s^{-1}

The frequency of this light is 6.62\times 10^{14}s^{-1}

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Consider the following reaction of calcium hydride (CaH2) with molten sodium metal: CaH2(s) 2 Na(l) -> 2 NaH(s) Ca(l) Identif
pashok25 [27]

Answer:

None of the species in the equation have undergone either oxidation or reduction

Explanation:

The easiest way to see if oxidation or reduction has happened is to compare oxidation numbers of each and every species before and after the reaction.

Calcium is two before the reaction and two after the reaction

Hydrogen is -1 before the reaction and -1 after the reaction

Sodium is one before the reaction and one after the reaction

Iodine is -1 before the reaction and -1 after the reaction.

For an oxidation to happen an increase in oxidation number has to happen.

For a reduction to happen, a decrease in oxidation number has to happen. None have happened

8 0
3 years ago
How many moles of o are in 15.0 mol fe(no3)3?
krok68 [10]
N[Fe(NO₃)₃]=15.0 mol

n(O)=3×3×n[Fe(NO₃)₃]=9n[Fe(NO₃)₃]

n(O)=9×15.0=135.0 mol
7 0
4 years ago
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Acetone used for rinsing glassware Group of answer choices
Marysya12 [62]

Answer: 2

Explanation: n/a

5 0
3 years ago
Calculate the ph of a 0.005 m solution of potassium oxide k2o
Alecsey [184]
First, we have to see how K2O behaves when it is dissolved in water:

K2O + H20 = 2 KOH

According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:

c(KOH) : c(K2O) = 1 : 2

So,

c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)

Now we can calculate pH:

pOH= -log c(OH⁻) = -log 0.01 = 2

pH= 14-2 = 12




3 0
3 years ago
g Using the complex based titration system: 50.00 mL 0.00250 M Ca2 titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i
kobusy [5.1K]

Answer:

Explanation:

a).

conc of Ca²⁺ =0.0025 M

pCa = -log(0.0025) = 2.6

logK,= 10.65 So lc = 4.47 x 10.

Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is  \alpha_{Y^{-4}} =0.81

So the Conditional Formation constant= K_f =0.81x 4.47 x10¹⁰

=3.62x10¹⁰

b)

At Equivalence point:

Ca²⁺ forms 1:1 complex with EDTA At equivalence point,

Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol

Number of moles of EDTA= 0.125 mol

Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL  

V e= 25.00 mL  

At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.  

[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M

                                                            {K^'}_f

                       Ca²⁺      +      Y⁴          ⇄     CaY²⁻

Initial                 0                  0                      0.001667

change             +x                  +x                     -x

equilibrium        x                    x                    0.001667 - x

{K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\

x^2=\frac{0.001667}{3.62\times10^{10}}=4.61\exp-14

x = 2.15×10⁻⁷

[Ca+2] = 2.15x10⁻⁷ M  

pca = —log(2 15x101= 6.7

3 0
4 years ago
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