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2 years ago

From the Bohr equation in the introduction, the calculated energy of an electron in the sixth Bohr orbit of a hydrogen atom is

Chemistry

1 answer:

Natalka [10]2 years ago

4 0

Answer:

= - 0.38 eV

Explanation:

Using Bohr's equation for the energy of an electron in the nth orbital,

E = -13.6 $\frac{Z^{2} }{n^{2} }$

Where E = energy level in electron volt (eV)

Z = atomic number of atom

n = principal state

Given that n = 6

⇒ E = -13.6 × $\frac{1^{2} }{6^{2} }$

= - 0.38 eV

Hope this was helpful.

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Now imagine you have several of such dipoles, and place them regularly between the plates. For this part of the pre-lab, you can

Dmitry_Shevchenko [17]

Answer:

Dipole is two equal but opposite charges in one. As like charges repels and unlike charges attracts each other. If dipole are in electric field region, it positive side will be attracted to the negative field and vis-versa. This will result to different movement or stationary based on the orientation of the field. If the interaction between other dipole is considered this effect may be obstructed

Explanation:

6 0

2 years ago

15 g of gold and 25 g of silver are mixed to form a single-phase ideal solid solution.

Tatiana [17]

Answer:

1. How many moles of solution are there. Ans: 0.3079193mol

2. Mole fraction for gold : 0.2473212

Mole fraction for silver: 0.7526787

3. Molar entropy of mixing for gold: 2.87285j/k

Molar entropy of mixing for silver: 1.77804j/k

4. Total entropy of mixing: 4.65089j/k

5. Molar free energy: -2325.445kj

6. Chemical potential for silver: -1750.31129j/mol

Chemical potential for gold: -575.13185j/mol

Explanation:

(1)

molar mass of silver = 107.8682g/mol

Molar mass of gold= 196.96657g/mol

Therefore mole = mass/molar mass

For silver: 25g/107.8682g/mol = 0.2317643mol

For gold: 15g/196.96657g/mol= 0.076155mol

Total number of mole= 0.2317643+0.076155= 0.30791193mol

(2)

Mole fraction for silver= 0.2317643/0.3079193= 0.7526787

Mole fraction for gold=0.076155/0.3079193=0.2473212

(3)

The molar entropy mixing ∆Sm= -RXi×lnXi

R= gas constant= 8.3144598

Xi = mole fraction

For silver:

-8.3144598×0.7526787( ln0.7526787)= 1.77804j/k

For gold:

-8.3144598×0.2473212( ln0.2473212)= 2.87285k/j

(4)

Total entropy= 1.77804+2.87285=4.65089k/j

(5)

Molar free energy change at 500°C

G=H-TS

Where G= Gibbs free energy

H= enthalpy,. T= Temperature, S= entropy

H=0, T=500+ 273=773k, S=4.65089

Therefore

G= 0- 773x4.65089= -3595.138kj

(6)

Chemical potential = Gibbs free energy × mole fraction

For silver:

-3595.138×0.7526787=-2705.9837j/mol

For gold:

-3595.138×0.2473212= -889.15381j/mol

8 0

2 years ago

Why would it have been wrong to write Li2SO4 as lithium sulfur oxide?

Bess [88]

Due to pyro-electric properties and molarity Li2SO4 cannot be written as lithium sulfur oxide.

Explanation:

Lithium sulfate is a white inorganic salt with the formula Li2SO4. It is the lithium salt of sulfuric acid.

Lithium sulfate has water solubility, though it does not follow the usual trend of solubility versus temperature — its solubility in water decreases with increasing temperature, as its dissolution is an exothermic process. This property is shared with few inorganic compounds, such as the lanthanoid sulfates.

Lithium sulfate has pyro-electric properties. When aqueous lithium sulfate is heated, the electrical conductivity also increases. The molarity of lithium sulfate also plays a role in the electrical conductivity optimal conductivity is achieved at 2M and then decreases.

Lithium sulfate has a rapid gastrointestinal absorption rate and complete following oral administration of tablets or the liquid form.

5 0

2 years ago

Can water (or any solvent) dissolve an infinite amount of solute?

anzhelika [568]

Answer:

b. is the answer hope this helps

8 0

2 years ago

Given the following unbalanced chemical reaction: As + NaOH Na3AsO3 + H2 What would be the coefficient of the NaOH molecule in t

barxatty [35]

Answer: -

Explanation: -

The given unbalanced chemical equation is As + NaOH -- > Na3AsO3 + H2

We see there 3 sodium on the right side from Na3AsO3.

But there are only 1 sodium on the left from NaOH.

So we multiply NaOH by 3.

As + 3 NaOH -- > Na3AsO3 + H2

Now we see the number of Hydrogen on the left is 3.

But the number of hydrogens is 2 on the left.

So, we multiply to get both sides 6 hydrogen.

As + 6NaOH -- > Na3AsO3 + 3 H2

Rebalancing for Na,

As + 6NaOH -- > 2Na3AsO3 + 3 H2.

Finally balancing As,

2 As + 6 NaOH -- > 2Na3AsO3 + 3H2

The coefficient of the NaOH molecule in the balanced reaction is thus 6

7 0

2 years ago