Answer:
the character developed in an individual as a result of environmental influence.
Explanation:
The equation for the reaction between NaOH and AlCl₃ is as follows;
3NaOH + AlCl₃ ---> 3NaCl + Al(OH)₃
the stoichiometry of NaOH : AlCl₃ is 3:1
3 moles of NaOH reacts with 1 mol of AlCl₃ to produce 1 mol of Al(OH)₃
the number of AlCl₃ moles reacted - 6.5 mol
molar mass of NaOH -(23 +16 +1) = 40 g/mol
the number of NaOH moles reacted = 57.0 g / 40 g/mol
NaOH moles = 1.425 mol
either NaOH or AlCl₃ is in excess and other is the limiting reactant.
limiting reactant is the reactant whose number of moles are fully consumed during the reaction. the reactant that is in excess will have leftover moles that are remaining after the reaction.
If AlCl₃ is the limiting reactant, number of NaOH moles would be thrice the amount of AlCl₃ present,
then number of NaOH moles that should be present - 6.5 * 3 = 19.5 mol
however there are only 1.425 mol of NaOH present, therefore AlCl₃ is in excess.
Then NaOH is the limiting reactant,
the amount of products formed depends on the amount of the limiting reactant present.
stoichiometry of NaOH : Al(OH)₃ is 3:1
the number of Al(OH)₃ moles produced = number of NaOH moles reacted / 3
number of Al(OH)₃ moles are - 1.425 mol /3 = 0.475 mol
molar mass of Al(OH)₃ = (27 +3*16 + 3*1) = 78 g/mol
mass of Al(OH)₃ produced = 78 g/mol * 0.475 mol = 37.05 g
1.94 * 6. 01 = 11.7 which is approximately same as 11 700. So the answer is 3).
Answer:
A. Because they are compounds, they cannot be pure substances.
Explanation:
The false statement from the given choices is that because they are compounds they cannot be pure substances. In fact, because they are compounds they are pure substances.
Pure substances are made up of elements and compounds and they have the following properties:
- All parts are the same throughout
- Composition is definite
- They cannot easily be separated or broken
- Separation by physical method is not easy
- They have unique sets of physical and chemical properties.
Answer:
The answer to your question is below
Explanation:
1)
Balanced chemical reaction
2CH₃OH + 3O₂ ⇒ 2 CO₂ + 4H₂O
Reactant Element Product
2 C 2
8 H 8
8 O 8
Molar mass of CH₃OH = 2[12 + 16 + 4]
= 2[32]
= 64 g
Molar mass of O₂ = 3[16 x 2] = 96 g
Theoretical proportion CH₃OH/O₂ = 64 g/96g = 0.67
Experimental proportion CH₃OH/O₂ = 60/48 = 1.25
Conclusion
The limiting reactant is O₂ because the Experimental proportion was higher than the theoretical proportion
2)
Balanced chemical reaction
S₈ + 12O₂ ⇒ 8SO₃
Reactant Elements Products
8 S 8
24 O 24
Molar mass of S₈ = 32 x 8 = 256 g
Molar mass of O₂ = 12 x 32 = 384 g
Theoretical proportion S₈ / O₂ = 256 / 384
= 0.67
Experimental proportion S₈ / O₂ = 40 / 35
= 1.14
Conclusion
The limiting reactant is O₂ because the experimental proportion was lower than the theoretical proportion.