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Blizzard
1 month ago
5

Ice fishermen sit on top of frozen lakes in the winter and catch fish in the liquid water below through holes cut in the ice she

et. The fact that the lake doesn't freeze solid is a good thing for those fish who live there and for the hungry fishermen. In this problem we will investigate how long it takes for this sheet of ice to form and why the lake doesn't freeze solid during the winter. The water on the surface of the lake only starts to freeze when the air temperature is below the freezing temperature of water.
Why does the ice only form on the surface and not throughout the entire volume of the lake?
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
8 0
This is due to the fact that below 4°c, water behaves differently than other substances and decreases in density as its temperature drops further.
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Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g
Softa [3030]

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

3 0
2 months ago
True or False: Molecules in a gas resist crowding and get as far apart as possible. Free electrons also resist crowding and get
ValentinkaMS [3465]

Answer:

This assertion is inaccurate.

Explanation:

The random nature of gas molecules results in their erratic motion and occasional collisions. While it is true that they tend to avoid being tightly packed, achieving the maximum separation from each other is not always feasible due to their lack of fixed positions. Consequently, gas molecules in a container cannot consistently maintain the furthest distance from their neighboring molecules.

In contrast, the separation among electrons is primarily influenced by repulsive forces, not random movement as in gases. Electrons maintain distance as a result of repulsion between similarly charged particles. Therefore, the arrangement of electrons on a charged copper sphere occurs not from a random distribution but rather due to repulsion, establishing a set distance between them.

4 0
2 months ago
A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t
inna [3103]

Answer:

The answer to the specified question will be "\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}".

Explanation:

Referring to the question,

\sum F_{x}

⇒  TCos \theta-F_{s}=0

⇒  T_{m}Cos \theta =F_{s}...(equation 1)

\sum F_{y}

⇒  TSin \theta+F_{N}=m_{g}

⇒  M_{g}-TSin \theta=F_{N}...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒  T_{m} Cos \theta = \mu_{s}F_{N}

By substituting the value of F_{N}, we derive

⇒  T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)

⇒  \mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}

4 0
2 months ago
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