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Alex_Xolod [135]
3 years ago
8

A piston compresses the air in the cylinder by doing 67.0 J of work on the air. As a result, the air gives off 149.0 J of heat t

o the surroundings. If the system is the air in the cylinder, what is the change in its internal energy?
Chemistry
1 answer:
harina [27]3 years ago
5 0

Explanation:

so sorry

don't know but please mark me as brainliest please

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Which has more atoms: 1 mole of hydrogen atoms or 1 mole of carbon-12 atoms?
Lyrx [107]
Each mole has 6.02 x10^23 atoms so they have same atoms
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3 years ago
B) Ethane has the formula C2H6. In terms of electronic structure,
k0ka [10]

Answer:

There is one single covalent bond between two carbon atoms.

Explanation:

We know that sharing of electrons form covalent bonds.

If we look upon K,L,M ,N shells of the carbon and hydrogen atoms.

We found that Hydrogen is having only  1 electron in K shell.

And Carbon on the other hand is having 2 electrons in K shell and 4 electrons in L shell.

So carbon have  4 valence electrons,and it can share 4 bonds with any relevant atom to complete its octet.

And Hydrogen requires  1 electron to complete its doublet.

Alkane general formula C_{n}H_{2n+2}

For ethane n=2

C_{2}H_{4+2} = C_{2}H_{6}

1 Carbon atom is shared by  3 Hydrogen.

The remaining one electron (4-3)=1 of carbon will be shared with another carbon atom.

An image of the sharing of electrons attached below,

Hence we have only  1 covalent bond between the two.

5 0
4 years ago
Read the following chemical equation.
Anton [14]

Answer:

Bromine gains an electron

Explanation:

According to oxidation and reduction

8 0
4 years ago
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What is the name of Fe(OH)3
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Answer:Iron(|||) Hydroxide

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4 0
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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