Hydrocarbons may vary from state to state depending on the length of the carbon chain. For methane to butane, they are usually in gas form. Starting from pentane, they are in the liquid form. For very long carbon chains, that occur as solids. Now, it depends if the pentane is in a nonpolar liquid or polar liquid. Since pentane is nonpolar, it is miscible in the liquid solvent. The movements would most likely be free flowing. But if he solvent is polar, the molecules would repel with one another. In the end, it will form two liquid phases, on for the pentane and one for the polar solvent.
Answer:
The number of copper atoms 12.405 ×10²³ atoms.
The number of silver atoms 13.13 ×10²³ atoms.
Beaker B have large number of atoms.
Explanation:
Given data:
In beaker A
Number of moles of copper = 2.06 mol
Number of atoms of copper = ?
In beaker B
Mass of silver = 222 g
Number of atoms of silver = ?
Solution:
For beaker A.
we will solve this problem by using Avogadro number.
The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.
While we have to find the copper atoms in 2.06 moles.
So,
63.546 g = 1 mole = 6.022×10²³ atoms
For 2.06 moles.
2.06 × 6.022×10²³ atoms
The number of copper atoms 12.405 ×10²³ atoms.
For beaker B:
107.87 g = 1 mole = 6.022×10²³ atoms
For 222 g
222 g / 101.87 g/mol = 2.18 moles
2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms
Reactions of Ethyl-3-pentenoate with all given reagents are given below.
Reaction with H₂ / Pd:
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.
Reaction with NaBH₄: Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.
Reaction with LiAlH₄: Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.
Result: The correct answer is
Option-A (Highlighted RED below).
Mass of lead (II) chromate is 51 g. The molecular formula is
and its molar mass is 323.2 g/mol
Number of moles can be calculated using the following formula:

Here, m is mass and M is molar mass.
Putting the values,

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.
84.34 grams of grams of iron (III) chloride that can be produced is maximum because Fe is the limiting reagent in this reaction and chlorine gas is excess reagent.
Explanation:
Balanced chemical equation:
2 Fe + 3 Cl2 → 2 FeCl3
DATA GIVEN:
iron = atoms
mass of chlorine gas = 67.2 liters
mass of FeCl3 = ?
number of moles of iron will be calculated as
number of moles = 
number of moles = 
number of moles = 0.52 moles of iron
moles of chlorine gas
number of moles = 
Putting the values in the equation:
n =
(atomic mass of chlorine gas = 70.96 grams/mole)
= 947.01 moles
Fe is the limiting reagent so
2 moles of Fe gives 2 moles of FeCl3
0.52 moles of Fe will give
= 
0.52 moles of FeCl3 is formed.
to convert it into grams:
mass = n X atomic mass
= 0.52 x 162.2 (atomic mass of FeCl3 is 162.2grams/mole)
<h3> = 84.34 grams </h3>