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Hoochie [10]
3 years ago
8

A sample of an ideal gas at 1.00 atm and a volume of 1.72 l was placed in a weighted balloon and dropped into the ocean. as the

sample descended, the water pressure compressed the balloon and reduced its volume. when the pressure had increased to 35.0 atm, what was the volume of the sample? assume that the temperature was held constant.
Chemistry
1 answer:
kolbaska11 [484]3 years ago
7 0
If we analyze the situation analytically, there are situations or states. Then, we are also given with 2 values of pressure and 1 value of volume. Lastly, temperature was set as constant. Thus, this means we use the Boyle's Law.

P₁V₁ = P₂V₂
Let's find V₂.
(1 atm)(1.72 L) = (35 atm)(V₂)
Solving for V₂,
<em>V₂ = 0.049 L</em>
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Answer:

b.

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When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The struc
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Answer:

The structure is shown below.

Explanation:

The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:

FC = V - (L + S/2)

Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.

The possible formulas, from the empiric one, are:

SCl, S₂Cl₂, and S₃Cl₃.

To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:

FC = 7 - (6 + 2/2) = 0

So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)

FC = 6 - (4+ 2/2) = +1

For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so

FC = 6 - (4 + 4/2) = 0

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6 0
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joja [24]

<u>Answer:</u> The concentration of CH_3COOH comes out to be 0.16 M.

<u>Explanation:</u>

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where,

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n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

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Hence, the concentration of CH_3COOH comes out to be 0.1862 M.

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