Answer : If a substance is the limiting reactant, then it limits the formation of products because in the reaction it is present in limited amount.
Explanation :
While observing a chemical reaction, we can tell about whether a reactant is limiting or excess.
Step 1 : first write the chemical reaction and then balanced the chemical equation.

Step 2 : convert the given masses into the moles if mass of
is 10.5 g and molar mass of
is 28 g/mole and the mass of hydrogen is 0.40 g and molar mass of hydrogen is 2 g/mole.


Step 3 : Now we have to determine the limiting reagent and excess reagent.

Now we conclude that
is the limiting reagent and hydrogen is an excess reagent.
Hypothesis :
Limiting reagent : It is the reagent in the chemical reaction that is totally consumed when the chemical reaction is complete. Limiting reagent limits the formation of products.
Answer:
Atom
Explanation:
The smallest unit which maintains an element's properties is an atom.
Answer:
3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂
Explanation:
3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂
In this type of reaction, one substance is replacing another:
A + BC → AC + B
In a single displacement reaction, atoms replace one another based on the activity series. Elements that are higher in the activity series. Also, if the element that is to replace the other in a compound is more reactive the reaction will occur. If it is less reactive, there will be no reation.
In the first equation, fluorine is more reactive than bromine. Therefore, bromine cannot replace bromine.
In the second equation, the displacement is between hydrogen and aluminium. Hydrogen is lower in the activity series, this implies that aluminum will replace it.
Answer:
d = 0.9 g/L
Explanation:
Given data:
Number of moles = 1 mol
Volume = 24.2 L
Temperature = 298 K
Pressure = 101.3 Kpa (101.3/101 = 1 atm)
Density of sample = ?
Solution:
PV = nRT (1)
n = number of moles
number of moles = mass/molar mass
n = m/M
Now we will put the n= m/M in equation 1.
PV = m/M RT (2)
d = m/v
PM = m/v RT ( by rearranging the equation 2)
PM = dRT
d = PM/RT
The molar mass of neon is = 20.1798 g/mol
d = 1 atm × 20.1798 g/mol / 0.0821 atm. L/mol.K × 273K
d = 20.1798 g/22.413 L
d = 0.9 g/L
D has a total of four significant figures.