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Ad libitum [116K]
3 years ago
10

Light is released when an excited electron moves from higher energy level back down to its lower, more stable energy level. The

resulting spectrum is a(n) _____ spectrum.
Chemistry
2 answers:
podryga [215]3 years ago
7 0
<span>The light spectrum that results from the light produced by an excited electron moving from a higher energy level back down to a lower energy level is an emission spectrum. The emission spectrum is formed by the electromagnetic radiation released by the transition in energy state.</span>
nignag [31]3 years ago
3 0
Answer is: emission spectrum.
When atom or molecule moves from a high energy level to a lower energy level than chemical element<span> or </span>chemical compound<span> emits spectrum of </span>frequencies<span> of </span>electromagnetic radiation with specific energy. That energy is <span>equal to the energy difference between the two levels.</span>



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Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
**<br> H₂ + O2-&gt;<br> Balancing equations from a reaction
goblinko [34]

Answer:

2H2+O2->2H2O

Explanation:

You have to balance the equation, you have 4 hydrogen and 2 oxygen atoms

8 0
2 years ago
A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.
kaheart [24]

The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L =  1.600M

In this case, we must set up the ICE table as shown;

              Br2(g) + I2(g) ↔ 2IBr(g)

I          0.6            1.6           0

C      -x                -x             +2x

E    0.6 - x         1.6 - x       1.190

If 2x = 1.190

x = 1.190/2

x = 0.595

The concentrations at equilibrium are;

[Br2] = 0.6 -  0.595 = 0.005

[I2] =   1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

Learn more about equilibrium constant:brainly.com/question/15118952

#SPJ1

4 0
1 year ago
Combustion of 6.90g of this compound produced 13.8g of CO2 and 5.64g of H2O. What is the empirical formula of the unknown compou
Alla [95]
The empirical formula for the unknown compound would be: C2H4O (2 molecules of Carbon, 4 molecules of Hydrogen, and 1 molecule of Oxygen)
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Steel is an alloy consisting of Fe with a small amount of C. Elemental Cr can be added to steel to make the steel less likely to
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The statement is true
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