Answer:
P₂ = 28.5 torr
Explanation:
Given data:
Initial pressure = 38 torr
Initial volume = 500 L
Final volume = 677 L
Final pressure = ?
Solution:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = Initial volume
P₂ = Final pressure
V₂ = Final volume
Now we will put the vales in formula.
P₁V₁ = P₂V₂
P₂ = P₁V₁ /V₂
P₂ = 38 torr × 500 L / 667 L
P₂ = 19000 torr. L / 667 L
P₂ = 28.5 torr
Answer:
CO2 < CH3Br < CH3OH < RbF
Explanation:
The boiling point of the compounds can be determined in terms of the strength of the intermolecular forces present in each compound.
Intermolecular forces are weak forces joining non-polar and polar molecules together. We have London dispersion forces, dipole-dipole forces of attraction, and hydrogen bonding.
London dispersion forces are weak attractions found between non-polar and polar symmetrical molecules. They are the weakest of all the electrical forces and they act between atoms and molecules e.g CO2
The dipole-dipole attractions are forces of attraction existing between polar unsymmetrical molecules. The dipole-dipole force of attractions is much stronger than London dispersion forces but weaker than Hydrogen bonding. e.g CH3Br
Hydrogen bonding is a special dipole-dipole attraction between polar molecules in which a hydrogen atom is directly joined to a highly electronegative atom (e.g. oxygen, nitrogen, or fluorine). Here, the bond in CH3OH is a hydrogen bond.
Ionic bonding is a bond that is formed between two kinds of atoms having a large electronegative difference such as in RbF,
Thus, in increasing order of boiling point;
CO2 < CH3Br < CH3OH < RbF
<span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span><span>Al
Cu
F
Co</span><span>
</span></span></span></span></span></span></span></span></span></span></span></span></span></span>
Answer:Step 1: Gather the Needed Materials (coffee Filter, Pencil, Beaker, Water Soluble Pen/marker, Water) Step 2: Cut a Strip of Filter Paper From the Coffee Filter and Using the Marker Draw a Thick Line Near the Bottom of the Filter Paper- About 1/4 Inch From the Bottom. Step 3: Allow the Water to Move Up the Paper for Approx. 5 Minutes and Then Remove the Strip From the Water Step 4: Students Should Identify the Individual Colors in the Chromatogram That Make Up the Markers Unique Mix of Ink. Step 5: Once Students Have Mastered the Technique Its Time to Put Their Skills to the Test in "The Case of the Celebrity Dog-napping" Step 7: Write the Ransom Note on a Piece of Filter Paper in One of the Suspect Inks.
Explanation: Chromatography is the physical separation of a mixture into its individual components. This lesson integrates mixtures, solutions & solubility into an activity where students will identify the solute & the solvent. Because chromatography can be used to separate the components of inks & dyes, students are able to identify the unique "recipe" that makes up the mixture of a specific brand of pen. After the students are taught the ink chromatography technique, the fun can continue with a mini forensic investigation, "The Case of the Celebrity Dog-napping".
hopes this helps
<u>Answer:</u> The correct answer is Option C.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of methane = 48.2 g
Molar mass of methane = 16 g/mol
Putting values in above equation, we get:
According to mole concept:
1 mole of compound contains 
number of molecules.
So, 3.0125 moles of methane will contain = 
number of molecules.
Hence, the correct answer is Option C.
