Answer:
-2092 kJ
Explanation:
Let's consider the chemical reaction that causes chromium to corrode in air.
4 Cr + 3 O₂ → 2 Cr₂O₃
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ΔH° - T × ΔS°
where,
- ΔH°: standard enthalpy of the reaction
- ΔS°: standard entropy of the reaction
ΔG° = -2256 kJ - 298 K × (-0.5491 kJ/K)
ΔG° = -2092 kJ
The composite material is composed of carbon fiber and epoxy resins. Now, density is an intensive unit. So, to approach this problem, let's assume there is 1 gram of composite material. Thus, mass carbon + mass epoxy = 1 g.
Volume of composite material = 1 g / 1.615 g/cm³ = 0.619 cm³
Volume of carbon fibers = x g / 1.74 g/cm³
Volume of epoxy resin = (1 - x) g / 1.21 g/cm³
a.) V of composite = V of carbon fibers + V of epoxy resin
0.619 = x/1.74 + (1-x)/1.21
Solve for x,
x = 0.824 g carbon fibers
1-x = 0.176 g epoxy resins
Vol % of carbon fibers = [(0.824/1.74) ÷ 0.619]*100 =<em> 76.5%</em>
b.) Weight % of epoxy = 0.176 g epoxy/1 g composite * 100 = <em>17.6%</em>
Weight % of carbon fibers = 0.824 g carbon/1 g composite * 100 = <em>82.4%</em>
Answer: 178.9 g
Explanation:
Density = 
find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³
1.0 cm³ = 1.0 mL
so 195.112 cm³ = 195.112 mL
plug value into density equation:
0.917 g/mL = (mass) / (195.112 mL)
and solve for mass!
Answer:

Explanation:
Given:
For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.
To find: fraction of the whole athletic field reserved for each fifth class
Solution:
Fraction of the whole athletic field reserved for four fifth classes = 
So, fraction of the whole athletic field reserved for each fifth class = 
Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.
To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.