Answer:
The mass of the lead
Explanation:
The 5g mass of the lead is an extensive property.
An extensive property is a physical property of matter which depends on the amount of matter that is present there in. Mass, volume e.t.c are all extensive properties. The more the quantity of the lead, the more its mass and the volume it occupies.
Melting point, boiling point, density are all intensive propeties. These properties do not rely on the amount of matter present. Any amount of lead will have the same density.
Answer:
0.800 mol
Explanation:
We have the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with moles of the compounds involved.
Step 1. <em>Gather all the information</em> in one place.
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 4.00 4.00
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Step 2. Identify the <em>limiting reactant
</em>
Calculate the <em>moles of CO₂</em> we can obtain from each reactant.
<em>From C₃H₈:</em>
The molar ratio of CO₂: C₃H₈ is 3:1
Moles of CO₂ = 4.00 × 3/1
Moles of CO₂ = 12.0 mol CO₂
<em>From O₂</em>:
The molar ratio of CO₂: O₂ is 3:5.
Moles of CO₂ = 4.00 × ⅗
Moles of CO₂ = 2.40 mol CO₂
O₂ is the limiting reactant because it gives the smaller amount of CO₂.
==============
Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.
The molar ratio of C₃H₈:O₂ is 1:5.
Moles of C₃H₈ = 4.00 × ⅕
Moles of C₃H₈ = 0.800 mol C₃H₈
Answer:
B. N2
Explanation:
The triple bonds pull the atoms closer together, and since N2 is the only molecule with the triple bond, it is the shortest bond length.
Answer:
6.4 L
Explanation:
When all other variables are held constant, you can use Boyle's Law to find the missing volume:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the theoretical volume by plugging the given values into the equation and simplifying.
P₁ = 3.2 atm P₂ = 1.0 atm
V₁ = 2.0 L V₂ = ? L
P₁V₁ = P₂V₂ <----- Boyle's Law
(3.2 atm)(2.0 L) = (1.0 atm)V₂ <----- Insert values
6.4 = (1.0 atm)V₂ <----- Simplify left side
6.4 = V₂ <----- Divide both sides by 1.0
The element iodine (I) is important for the fast and hastened metamorphosis of frog-tadpoles. As amphibians, the tadpoles can live in water and land but when they are born they are iodine-deficient. Tadpoles that do not receive ample amount of iodine become tadpoles until the end of their days.
