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trasher [3.6K]
3 years ago
7

g line a) The emission line with the shortest wavelength. b) The absorption line with the shortest wavelength. c) The emission l

ine with the lowest energy. d) The absorption line with the lowest energy. e) The emission line with the lowest frequency. f) The line corresponding to the ionization energy of hydrogen.
Chemistry
1 answer:
NISA [10]3 years ago
3 0

Answer:

a) 4

b) 2

c) 5

d) 2

e) 4

f) 3

Explanation:

The complete question is shown in the image attached to this answer.

We know that the shorter the wavelength, the greater the energy. The greatest energy and shortest wavelength among the emission lines is 4.

Applying the same argument as above, the greatest energy catapults the electron from energy level n=1 to n=4. This will corresponds to the shortest wavelength since energy is inversely proportional to wavelength.  

Frequency is also inversely related to wavelength for two waves travelling at the same speed. Hence, the transition that corresponds to the highest energy and shortest wavelength in the emission spectrum will have the highest frequency.

The emission line with the lowest energy occurs when the electron moves from n=2 to n=1.

The ionization energy is the energy required to remove an electron from the hydrogen atom. It corresponds to the transition n=1 to n=∞

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The answer is (3)
 I always use the acronym SNAP
S- symmetrical is
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7 0
3 years ago
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Which is true about reaction rates?
Oksana_A [137]

Answer:

the correct choice would be: 4. Reaction rates are not related to spontaneity.

Explanation:

The speed of reaction is independent of spontaneity, since it depends on other factors such as the amount of reagent.

The amount of reagent is one of the variables that most influences the speed of reaction, the greater the amount of reagent, the faster the reaction will be and thus the speed will be increased.

3 0
4 years ago
In the first 15.0 s of the reaction, 1.9×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.480 L . What is the
N76 [4]

The question is incomplete, here is the complete question:

Consider the following reaction:  2N_2O(g)\rightarrow 2N_2(g)+O_2(g)

In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?

<u>Answer:</u> The average rate of appearance of oxygen gas is 2.64\times 10^{-3}M/s

<u>Explanation:</u>

We are given:

Moles of oxygen gas = 1.9\times 10^{-2}moles

Volume of solution = 0.480 L

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

So, \text{Molarity of }O_2=\frac{1.9\times 10^{-2}mol}{0.480L}=0.0396M

The given chemical reaction follows:

2N_2O(g)\rightarrow 2N_2(g)+O_2(g)

The average rate of the reaction for appearance of O_2 is given as:

\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}

Or,

\text{Average rate of appearance of }O_2=\frac{C_2-C_1}{t_2-t_1}

where,

C_2 = final concentration of oxygen gas = 0.0396 M

C_1 = initial concentration of oxygen gas = 0 M

t_2 = final time = 15.0 s

t_1 = initial time = 0 s

Putting values in above equation, we get:

\text{Average rate of appearance of }O_2=\frac{0.0396-0}{15-0}\\\\\text{Average rate of appearance of }O_2=2.64\times 10^{-3}M/s

Hence, the average rate of appearance of oxygen gas is 2.64\times 10^{-3}M/s

8 0
4 years ago
How many moles of H2 would be contained in 4.0 L of the gas at 202.6 kPa and 127°C?
kicyunya [14]
Use PV =nRT. Rearrange it to n = PV/RT.
P = 202.6 kPa
V = 4.0L
R = 8.314 kPa*L/mol*K
T = 127 °C + 273 = 400 K
Plug it in and solve. I got 0.24 moles of H2. 
6 0
4 years ago
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Balance the following chemical equation: <br> ___HCO + ___O —&gt; ___H2 + ___CO3
yawa3891 [41]

Answer: 2HCO + 4O → H2 + 2CO3

Explanation: Oxomethyl + Oxygen = Dihydrogen + Carbon Trioxide

Reaction Type: SINGLE REPLACEMENT

***If you found my answer helpful, please give me the brainliest, please give a nice rating, and the thanks ( heart icon :) ***

7 0
3 years ago
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