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Nadusha1986 [10]
3 years ago
9

Do we generally break chemical bonds when we cool molecules?

Chemistry
1 answer:
olga_2 [115]3 years ago
7 0
The breaking of chemical bonds never releases energy to the external environment. Energy is only released when chemical bonds are formed. In general, a chemical reaction involves two steps: 1) the original chemical bonds between the atoms are broken, and 2) new bonds are formed. These two steps are sometimes lumped into one event for simplicity, but they are really two separate events. For instance, when you burn methane (natural gas) in your stove, the methane is reacting with oxygen to form carbon dioxide and water.
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Is barium hydroxide Ba(OH)2 or is BaOH?
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Answer:

What is the charge on the barium ion and what is the charge of the hydroxide ion.

Explanation:

To get the correct formula they have to add to zero over all.

6 0
3 years ago
Differentiate between the three types of simple permanent tissues on the basis of their cell walls​
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Answer:

Parenchyma is the most simple and specialized tissue which is concerned mainly with the vegetative activities of the plant. The cells are isodiametric with well-developed intercellular spaces, vacuolated cytoplasm and cellulosic cell wall.

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3 years ago
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,

  • m(\text{Hg}) = m(\text{residure})
  • 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}.

8 0
3 years ago
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The correct answer is slow
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