Answer:
H₂SO₄ will be the limiting reagent.
Explanation:
The balanced reaction is:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction).
You can use a simple rule of three as follows: if by stoichiometry 2 moles of Al(OH)₃ reacts with 3 moles of H₂SO₄, how much moles of H₂SO₄ will be needed if 0.4 moles of Al(OH)₃ react?

moles of H₂SO₄= 0.6 moles
But 0.6 moles of H₂SO₄ are not available, 0.40 moles are available. Since you have less moles than you need to react with 0.4 moles of Al(OH)₃, H₂SO₄ will be the limiting reagent.
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Answer:

Explanation:
From the question we are told that:
Moles of sample n=3.20_mol
Volume V=350mL
Temperature T=300k
Generally the equation for ideal gas is mathematically given by





Answer is: the average atomic mass 217.606 amu.
Ar₁= 203.973 amu; the average atomic mass of isotope.
Ar₂ = 205.9745 amu.
Ar₃ = 206.9745 amu.
Ar₄ = 207.9766 amu.
ω₁ = 1.40% = 0.014; mass percentage of isotope.
ω₂ = 24.10% = 0.241.
ω₃ = 22.10% = 0.221.
ω₄ = 57.40% = 0.574.
Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.
Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.
Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.
Ar = 217.606 amu.
But abundance of isotopes is greater than 100%.
It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.