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2 years ago

The characteristic bright–line spectrum of potassium is produced when electrons

Chemistry

1 answer:

OLEGan [10]2 years ago

8 0

Answer: B.

Explanation:

When the electron changes levels, it decreases energy and the atom emits photons. The photon is emitted with the electron moving from a higher energy level to a lower energy level. The energy of the photon is the exact energy that is lost by the electron moving to its lower energy level.

Hope this helps!

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Classify each process by its individual effect on the entropy of the universe, s. hydrogen peroxide slowly decomposing into wate

Jobisdone [24]

Hydrogen Peroxide slowly decomposing into water and oxygen increases the entropy of the universe. Any reaction of a system always increases the degree of the orderliness of the universe. Decomposition is one of the best examples of increasing the entropy. This is when substance is broken down further to elements or other compounds.

4 0

2 years ago

What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B

natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

$A = A_{0}e^{-kt}$

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

$ln[A] = ln[A]_{0}-kt$

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

7 0

3 years ago

has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a

Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

$A + B \rightarrow C$

Initial : 0.3 0.4 0

Change: -x -x x

Equilbm: (0.3 - x) (0.4 - x) x

We know that, relation between standard free energy and equilibrium constant is as follows.

$\Delta G = -RT ln K$

Putting the given values into the above formula as follows.

$\Delta G = -RT ln K$

$-3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})$

x = 0.1417

Hence, at equilibrium

[A] = 0.3 - 0.1417

= 0.1583 M

[B] = 0.4 - 0.1417

= 0.2583 M

[C] = 0.1417 M

5 0

2 years ago

Alveoli are tiny sacs of air in the lungs. Their average diameter is 4.50 × 10−5 m. Calculate the uncertainty in the velocity of

Arisa [49]

<u>Answer:</u> The uncertainty in the velocity of oxygen molecule is $4.424\times 10^{-5}m/s$

<u>Explanation:</u>

The diameter of the molecule will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $4.50\times 10^{-5}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of oxygen molecule = $5.30\times 10^{-26}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg}=4.424\times 10^{-5}m/s$

Hence, the uncertainty in the velocity of oxygen molecule is $4.424\times 10^{-5}m/s$

8 0

2 years ago

I need some help with this. I’m so confused on what to do and an explanation to what the answer is and why would be ideal.

Anvisha [2.4K]

Answer:

who do I submit my questions to there is no submit or done sign I download this app for help not to waste my money and time. I need help with math please don't let us start off on the wrong foot

8 0

2 years ago