<span>We need to calculate the equivalent amount in units of moles of ammonium ions from the mass units. For this we need the molar mass of the substances involved. We calculate as follows:
31.3 g </span>(NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol <span>ammonium ions</span>
Because if you change two things then you do not know which one has affected or altered the dependent variable. if you only change one then you know what exactly changed and why
Composer and performer of Old Occitan lyric poetryduring the High Middle Ages (1100–1350). Since the word troubadour is etymologically masculine, a female troubadour is usually called a trobairitz.
Explanation:
The given data is as follows.
Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml
- Therefore, molarity of toulene is calculated as follows.
Molarity = 
= 
= 4.11 M
Hence, molarity of toulene is 4.11 M.
- As molality is the number of moles of solute present in kg of solvent.
So, we will calculate the molality of toulene as follows.
Molality = 
= 
= 8.6 m
Hence, molality of given toulene solution is 8.6 m.
- Now, calculate the number of moles of toulene as follows.
No. of moles = 
= 
= 0.8227 mol
Now, no. of moles of benzene will be as follows.
No. of moles = 
= 
= 1.2239 mol
Hence, the mole fraction of toulene is as follows.
Mole fraction = 
= 
= 0.402
Hence, mole fraction of toulene is 0.402.
- As density of given solution is 0.857
so, we will calculate the mass of solution as follows.
Density = 
0.857
=
(As 1
= 1 g)
mass = 171.4 g
Therefore, calculate the mass percent of toulene as follows.
Mass % = 
= 
= 44.22%
Therefore, mass percent of toulene is 44.22%.
Answer:
2Li + F₂ → 2LiF
Explanation:
The reaction expression is given as:
Li + F₂ → LiF
We are to balance the expression. In that case, the number of atoms on both sides of the expression must be the same.
Let use a mathematical approach to solve this problem;
Assign variables a,b and c as the coefficients that will balance the expression:
aLi + bF₂ → cLiF
Conserving Li: a = c
F: 2b = c
let a = 1, c = 1 and b =
Multiply through by 2;
a = 2, b = 1 and c = 2
2Li + F₂ → 2LiF