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ddd [48]
3 years ago
8

Determine the molar mass of a compound that has a density of 0.1633 g/L at STP. (show work)

Chemistry
1 answer:
hodyreva [135]3 years ago
6 0

Answer:

                     M.Mass  =  3.66 g/mol

Data Given:

                  M.Mass  =  M = ??

                  Density  =  d  =  0.1633 g/L

                  Temperature  =  T  =  273.15 K (Standard)

                  Pressure  =  P  =  1 atm (standard)

Solution:

              Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.

                                             P V = n R T    ---- (1)

Also, we know that;

                       Moles  =  n  =  mass / M.Mass

Or,                                   n  =  m / M

Substituting n in Eq. 1.

                                             P V = m/M R T   --- (2)

Rearranging Eq.2 i.e.

                                             P M = m/V R T   --- (3)

As,

                     Mass / Volume = m/V = Density = d

So, Eq. 3 can be written as,

                                             P M = d R T

Solving for M.Mass i.e.

                                             M = d R T / P

Putting values,

M  =  0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm

M  =  3.66 g/mol

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A. absorbs light

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In ____________ reactions, certain molecules are broken down; in the process they release energy which is captured when high-ene
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Exergonic ,Endergonic,low concentration area,high

Explanation:

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Select the statements that correctly describe a buffer. a) An acid added to the buffer solution reacts with the weak base of the
Tom [10]

Answer:

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Explanation:

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Answer:

see explanation below

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You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.

According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:

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Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:

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