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Gekata [30.6K]
1 year ago
14

The energy difference between the 5d and 6s sublevels in gold accounts for its color. Assuming this energy difference is about 2

.7 eV (electron volt; 1 eV = 1.602 x 10⁻¹⁹ J), explain why gold has a warm yellow color.
Chemistry
1 answer:
sergejj [24]1 year ago
7 0

Answer:

\lambda=459.1\times 10^{-7}\ m = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

Explanation:

Given that:- Energy = 2.7 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.602 × 10⁻¹⁹ J

So, Energy = 2.7\times 1.602\times 10^{-19}\ J=4.33\times 10^{-19}\ J

Considering:-

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

Where,  

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

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

\lambda is the wavelength of the light

So,  

4.33\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

4.33\times \:10^{26}\times \lambda=1.99\times 10^{20}

\lambda=459.1\times 10^{-7}\ m = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

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Calculate the mass defect of the boron nucleus 11 5b. The mass of neutral 11 5b is equal to 11.009305 atomic mass units.
Vlad1618 [11]

Answer:

0.07906687 amu

Explanation:

For Boron ₅B¹¹, the number of protons is 5 and the mass is 11. The mass is the number of protons plus the number of neutrons, so:

neutrons = 11 - 5 = 6

The mass of an atom is concentrated in the nucleus, so it is the mass of the protons + the mass of the neutrons. The mass of 1 proton is 1.00727647 amu/proton, and the mass of 1 neutron: 1.00866492 amu/neutron, so for the element given the theoretical mass (mt) is:

mt = 5* 1.00727647 amu/proton + 6*1.00866492 amu/neutron

mt = 11.08837187 amu

The mass defect (md) is the theorical mass less the real mass:

md = 11.08837187 - 11.009305

md = 0.07906687 amu

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2 years ago
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I NEED HELP ASAP, WILL MARK BRAINLEST!
Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial        SiCl₄                   O₂                    SiO₂

 1           120 g                  240 g              38.2 g

 2           75 g                   50 g                25.2 g

<u>Percentage yield for trial 1</u>

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

 molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

 120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

 For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

 .706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

 molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

 .706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

 actual/theoretical x 100

 38.2 / 42.44 = .900 = <u>90.0% yield</u>

<u>Leftover reactant for trial 1</u>

We know oxygen is the excess reactant.

Mass to moles:

 molar mass O₂ = 32.00 g/mol

 240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

 7.5 - .706 = 6.8 moles left over

Moles to mass:

 6.8 mol O₂ x 32.00g/mol =<u> 217.4 g O₂</u>

<u />

<u>Percentage yield for trial 2</u>

Mass to moles:

 molar mass SiCl₄: 169.9 g/mol

 75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

 For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

 .441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

 molar mass SiO₂: 60.09 g/mol

 .441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

 actual/theoretical x 100

 25.2 / 26.5 = .950 = <u>95.0% yield</u>

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

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2 years ago
If radioactive caesium was reacted with chlorine, would you expect the caesium chloride produced to be radioactive? Explain you
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Yes due to the radioactivity having nothing to do with the chemical equation given it will release radiation at a rate determined by it's half life.
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A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both s
marusya05 [52]

Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

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The mole fraction of species 1 in vapor phase is 0.9

4 0
1 year ago
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Saccharin is considered as weak acid:
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3 0
1 year ago
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