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Stella [2.4K]
3 years ago
13

The unique properties of water enable life to exist on Earth. Which of these is a property of pure water?

Chemistry
1 answer:
Lera25 [3.4K]3 years ago
3 0

Answer:

Capillarity , Adhesion and cohesion . The unique property of water enable life to exist . The most important property of water is movement of materials due to water , maintenance and growth and reproduction .

Explanation:

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Compare urban and rural land .how are each used?
Ksivusya [100]

Answer:

"Urban area is used for the buildings, industries etc, whereas the rural areas are used for cultivation, forest cover etc."

Explanation:

When we talk about the urban land they are mostly covered with buildings, industries, roads and apartments and also the municipal structures. But when we talk about the rural land it has less dense population with forest cover, agricultural lands, rangeland and also different land cover type. In the urban areas the population density is very high. Those areas that located in the outskirt of the town are the rural areas.

5 0
2 years ago
Newton’s second law of motion is F=ma A net force of 60 N north acts on an object with a of 30 kg. Use Newton’s second law of mo
horsena [70]
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8 0
3 years ago
Read 2 more answers
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

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  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

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3 years ago
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Answer:

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Explanation:

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Explanation:

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