C. Fixing a wineglass requires a large decrease in entropy.
Answer:
The time required to melt the frost is 3.25 hours.
Explanation:
The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .
The heat transferred per unit area can be expressed as:

being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).

If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume

The mass of the frost can be estimated as

Then, the amount of heat needed to melt this surface (1 m²) of frost is

The time needed to melt the frost can be calculated as

.answer:
so the first one is analogous
the second is homologous
and the third is analogous
explanation:
analogous. this means they share a similar function (flight) but do not have the same embryonic origin.
Answer:
- 602 mg of CO₂ and 94.8 mg of H₂O
Explanation:
The<em> yield</em> is measured by the amount of each product produced by the reaction.
The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.
The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

To calculate the yield follow these steps:
<u>1. Mole ratio</u>

<u />
<u>2. Convert 175mg of fluorene to number of moles</u>
- Number of moles = mass in grams / molar mass
<u>3. Set a proportion for each product of the reaction</u>
a) <u>For CO₂</u>
i) number of moles


ii) mass in grams
The molar mass of CO₂ is 44.01g/mol
- mass = number of moles × molar mass
- mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg
b) <u>For H₂O</u>
i) number of moles

ii) mass in grams
The molar mass of H₂O is 18.015g/mol
- mass = number of moles × molar mass
- mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
As you get higher the atmospheric pressure lowers. The pressure in the packet of crisps has the pressure at which it has been closed (pressure at the surface of the earth). This means that the air molecules in the packet press harder outside than the air molecules in the atmosphere press on the packet.