Answer:
Explanation:
Problem 1
<u>1. Data</u>
<u />
a) P₁ = 3.25atm
b) V₁ = 755mL
c) P₂ = ?
d) V₂ = 1325 mL
r) T = 65ºC
<u>2. Formula</u>
Since the temeperature is constant you can use Boyle's law for idial gases:

<u>3. Solution</u>
Solve, substitute and compute:


Problem 2
<u>1. Data</u>
<u />
a) V₁ = 125 mL
b) P₁ = 548mmHg
c) P₁ = 625mmHg
d) V₂ = ?
<u>2. Formula</u>
You assume that the temperature does not change, and then can use Boyl'es law again.

<u>3. Solution</u>
This time, solve for V₂:

Substitute and compute:

You must round to 3 significant figures:

Problem 3
<u>1. Data</u>
<u />
a) V₁ = 285mL
b) T₁ = 25ºC
c) V₂ = ?
d) T₂ = 35ºC
<u>2. Formula</u>
At constant pressure, Charle's law states that volume and temperature are inversely related:

The temperatures must be in absolute scale.
<u />
<u>3. Solution</u>
a) Convert the temperatures to kelvins:
- T₁ = 25 + 273.15K = 298.15K
- T₂ = 35 + 273.15K = 308.15K
b) Substitute in the formula, solve for V₂, and compute:

You must round to two significant figures: 290 ml
Problem 4
<u>1. Data</u>
<u />
a) P = 865mmHg
b) Convert to atm
<u>2. Formula</u>
You must use a conversion factor.
Divide both sides by 760 mmHg

<u />
<u>3. Solution</u>
Multiply 865 mmHg by the conversion factor:

Answer:
We are given:
Volume (V) = 0.25 L
Pressure (P) = 0.93 atm
Temperature (T) = 15.4°C OR 288.4 K
<u>Solving for the number of moles of CO₂:</u>
From the ideal gas equation:
PV = nRT
replacing the variables
0.93 * 0.25 = n (0.082)(288.4)
n = 0.00983 moles
<u>Number of molecules:</u>
Number of moles= 0.00983
number of molecules in 1 mole = 6.022 * 10²³
Number of molecules in 0.00983 moles = 0.00983 * 6.022 * 10²³
Number of molecules = 5.91 * 10²¹
Answer: The activation energy Ea for this reaction is 22689.8 J/mol
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
![ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
= rate constant at temperature
= 
= rate constant at temperature
=
= activation energy = ?
R= gas constant = 8.314 J/kmol
= temperature = 
= temperature = 
Putting in the values ::
![ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7B4.8%5Ctimes%2010%5E8%7D%7B2.3%5Ctimes%2010%5E8%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B649%7D%20-%20%5Cfrac%7B1%7D%7B553%7D%5D)

The activation energy Ea for this reaction is 22689.8 J/mol
It respresents the higher energy level than 627nm .
<h3>What is a emission line ? </h3>
Emission lines are the glowing hot gas emits lines of light whereas absorption line refers to the tendency of cool atmospheric gas to absorb the same line of light.Some lights produce dark band when the light passes through gas in the atmosphere . There are two line spectrum and absorption.
spectrum is an excitement of electrons from lower to higher energy levels and when it comes back it releases energy in the terms of colourful lights .
It represents the higher energy levels than 627nm because Energy is inversly proportional to wavelength .
to learn more about Emission lines click here
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D only..
A is pure compound
B is mixture of compound and element
C is mixture of compounds
D is mixture of elements