Answer:
The object at 50°C will have a higher kinetic energy.
Explanation:
Temperature is a measure of the average kinetic energy of the particles in an object. As you introduce more energy into the system (e.g. heat the object), the particles on average move faster because they have more kinetic energy.
Heterogeneous-a mixture of two or more things
solution-a substance that dissolves
suspension-mixture where solid particles do not dissolve
colloid-does not settle and cannot be seperated
Heterogeneous
- tuna casserole
-chocolate chip cookies
-gelatin dessert
-green salad
Solution
-oil (?) depends
Suspension
-vinegar salad dressing
-oil (?) it could go in either catagory
Colloid
-vinegar salad dressing
-oil (?)
Answer:
The specific rotation of D is 11.60° mL/g dm
Explanation:
Given that:
The path length (l) = 1 dm
Observed rotation (∝) = + 0.27°
Molarity = 0.175 M
Molar mass = 133.0 g/mol
Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol
Concentration in (g/mL) = 23.275 g/L
Since 1 L = 1000 mL
Concentration in (g/mL) = 0.023275 g/mL
The specific rotation [∝] = ∝/(1×c)
= 0.27°/( 1 dm × 0.023275 g/mL
)
= 11.60° mL/g dm
Thus, the specific rotation of D is 11.60° mL/g dm
The sub-atomic particles of an atom are the proton, electron and the neutron. An electron has a charge of -1 and a smaller
mass than a proton. Proton has the same mass with the neutron. The ratio
between the mass of a proton and an electron is about 2000. An electron has an
equal value but negative charge with the proton.
Answer:
3224 kJ/mol
Explanation:
The combustion of benzoic acid occurs as follows:
C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE
The change in temperature in the reaction is the change due the energy released, that is:
3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts
To find the heat released per mole we have to find the moles of benzoic acid:
<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>
1.250g * (1mol / 122.12g) = 0.0102 moles
<em />
The dE combustion per mole of benzoic acid is:
33.00kJ / 0.0102moles =
<em>3224 kJ/mol </em>
