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oksano4ka [1.4K]
3 years ago
11

Why do you think the process to create a protected area takes so long?

Chemistry
1 answer:
Paladinen [302]3 years ago
4 0

The process to create a protected area takes a long time because there's numerous things that have to be taken in account and sorted out before the area gets that kind of legal status.

Usually there's people living in the area that is supposed to be protected, though it is mostly small settlements. The people living there are connected with that area, use the resources, and their lives are largely based and dependent on it. In order for any problems to be avoided, these people should all be educated, deals to be made with them, and properly negotiated so that the area wouldn't have any damage, but also them to be able to sustain themselves.

The boundaries of the area are also complicated, it it takes time to sort out what parts should be included and what not. Basically to set the priorities.

People should be employed to protect and take care of the area, so they have to undergo a training, education, and to get familiar with it.

The legalization of the area as such can sometimes too make problems and take some time.

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How many molecules of nitrogen gas are in 15 L of nitrogen gas
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Answer:

3.98\times 10^{23}

Explanation:

Since we have $N_A$ number of molecules in 22.4 L, in 15 L we have:

$\frac{15}{22.4}\times 6.023 \times 10^{23}$ molecules.

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Why ice floats on water is a physical change​
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4 years ago
What three achievements in space exploration that involved The United States
Marta_Voda [28]
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4 years ago
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What is an ideal gas? how many ideal gases are there in the universe?
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6 0
4 years ago
Camphor (C10H16O) melts at 179.8°C, and it has a particularly large freezing-point-depression constant, Kf= 40.0ºC/m. When 0.186
Usimov [2.4K]

Answer:

Molar mass for the unknown solute is 109 g/mol

Explanation:

Freezing point depression is the colligative property that must be applied to solve the question.

T°F pure solvent - T°F solution = Kf . m

Let's analyse the data given

Camphor → solvent

Unknown solute → The mass we used is 0.186g

T°F pure solvent = 179.8°C and T°F solution = 176.7°C.

These data help us to determine the ΔT → 179.8°C - 176.7°C = 3.1°C

So we can replace → 3.1°C  = 40°C/m . m

m = 3.1°C / 40 m/°C → 0.0775 mol/kg

We have these moles of solute in 1kg of solvent, but our mass of camphor is 22.01 g (0.02201 kg).

We can determine the moles of solute → molality . kg

0.0775 mol/kg . 0.02201 kg = 1.70×10⁻³ moles

Molar mass → mass (g) / moles → 0.186 g / 1.70×10⁻³ mol = 109 g/mol

3 0
3 years ago
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