Answer:
0.218 M of Pb(NO3)2
Explanation:
Equation of the reaction
Pb(NO3)2(aq) + 2NaCl(aq) --> PbCl2(s) + 2NaNO3(aq)
1 mole of Pb(NO3)2 reacts to precipitate 1 mole of PbCl2
Molar mass of PbCl2 = 207 + (35.5*2)
= 278 g/mol
Number of moles of PbCl2 precipitated = mass/molar mass
= 12.11/278
= 0.04356 mol
Since 0.04356 moles of PbCl2 was precipitated, therefore by stoichiometry; 0.04356 moles of Pb(NO3)2 reacted.
Molarity is defined as the number of moles of solute in 1 liter of solution.
Molarity = number of moles/volumes
= 0.04356/0.2
= 0.218 M
Huh I do not understand this
Answer:
See explanation
Explanation:
The particular reactants in the Fischer esterification reaction were not stated.
Generally, a Fischer esterification is a reaction that proceeds as follows;
RCOOH + R'OH ⇄RCOOR' + H2O
This reaction occurs in the presence of an acid catalyst.
We can shift the equilibrium of this reaction towards the products side in two ways;
I) use of a large excess of either of the reactants
ii) removal of one of the products as it is formed.
Any of these methods shifts the equilibrium of the Fischer esterification reaction towards the products side.
N₂ : limiting reactant
H₂ : excess reactant
<h3>Further e
xplanation</h3>
Given
mass of N₂ = 100 g
mass of H₂ = 100 g
Required
Limiting reactant
Excess reactant
Solution
Reaction
<em>N₂+3H₂⇒2NH₃</em>
mol N₂(MW=28 g/mol) :

mol H₂(MW= 2 g/mol) :

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants
From the equation, mol ratio N₂ : H₂ = 1 : 3, so :

N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant
Thomson<span> is the scientist who designed an experiment that enabled the first successful detection of an individual subatomic particle. </span>J.J. Thomson<span> (Sir </span>Joseph John Thomson<span>, 1856-1940), who demonstrated in 1897 that "cathode rays" consisted of negatively-charged particles, later named electrons.</span>