Answer:
Percent error = 3.7%
Explanation:
Given data:
Density of Al cylinder = ?
Weight of cylinder = 18 g
Diameter = 1.3 cm
Height = 5.2 cm
Actual density of Al = 2.7 g/cm³
Percent error = ?
Solution:
First of all we will calculate the volume of cylinder through given formula.
V = πr²h
r = diameter /2
V = 22/7 × (0.65 cm)²× 5.2 cm
V = 22/7 × 0.4225cm²× 5.2 cm
V = 6.89 cm³
Now we will calculate the density.
d = m/v
d = 18 g/ 6.89 cm³
d = 2.6 g/cm³
Percent error:
Percent error = measured value - actual value /actual value × 100
Percent error = 2.6g/cm³ - 2.7g/cm³ /2.7g/cm³ × 100
Percent error = 3.7%
Negative sign shows that measured or experimental value is less than actual value.
An element is a pure substance
Answer: 1. Mass and type of solution
2. Type of material
3. Temperature
Answer: The MAJOR energetic contribution to this property is the self association that maximizes the entropy of the water molecules.
Explanation:
Hydrophobic molecules (like oil) tend to self-associate in water rather than dissolve in it. The MAJOR energetic contribution to this property is the self association of this oil which increases the degree of disorderliness of the water molecules.
Answer:
11.3 g of H₂O will be produced.
Explanation:
The combustion is:
2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O
First of all, we determine the moles of the reactants in order to find out the limiting reactant.
8 g / 114g/mol = 0.0701 moles of octane
37g / 32 g/mol = 1.15 moles of oxygen
The limiting reagent is the octane. Let's see it by this rule of three:
25 moles of oxygen react to 2 moles of octane so
1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.
We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:
2 moles of octane produce 18 moles of water
Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.
We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.
