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BlackZzzverrR [31]
3 years ago
5

Derive an expression similar to the equation for the energy levels above for the single electron in He+ and Li2+ . Calculate the

numerical values for the 1s levels (n = 1)?
Chemistry
1 answer:
VikaD [51]3 years ago
3 0

Answer:

Explanation:

from Bohr's equation,

E = -Z²R/n²

R = 13.6 eV

Z = atomic number of element

for 1s energy level n= 1

E = -(Z)² x (13.6)/(1)²

E = -13.6Z²

calculating the numerical for the 1s energy levels for He+ and Li2+

- for He+

E = -13.6 * (2)² = -54.4eV

- for Li2+

E = -13.6 * (3)² = -122.4eV

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6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/
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Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}

(b) Percent transmission

A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}

7) Absorbance

A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}

8 0
3 years ago
2.7g of aluminum metal is placed into solution hydrochloric acid to form aluminum chloride and hydrogen gas.
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2 years ago
Which of these statements is not true about chemical reaction rates?
xeze [42]
Temperature can change a reaction rate because adding or taking away heat means energy is being added or taken away. When energy is added, the particles speed up, so there is a greater chance of the reactants colliding to form the products, which increases the reaction rate. When energy is taken away, the particles more slower, so they don't collide as easily, which slows down the reaction rate.

Therefore, the answer is D.
3 0
2 years ago
Read 2 more answers
How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?
zloy xaker [14]

Answer:

m_{HCl}=36.1gHCl

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl

Regards!

4 0
2 years ago
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