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Artemon [7]
3 years ago
10

Name the following alkane molecule:

Chemistry
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

It's the A. 2-bromobutane

Explanation:

Have a good day

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What mass of NaCl is dissolved in 150 g of water in a .050<br> msolution?
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Answer:

0.4383 g

Explanation:

Molality is defined as the moles of the solute present in 1 kg of the solvent.

It is represented by 'm'.

Thus,  

Molality\ (m)=\frac {Moles\ of\ the\ solute}{Mass\ of\ the\ solvent\ (kg)}

Given that:

Mass of solvent, water = 150 g = 0.15 kg ( 1 g = 0.001 g )

Molality = 0.050 m

So,

0.050=\frac {Moles\ of\ the\ solute}{0.15}

Moles = 0.050\times 0.15\ mol= 0.0075\ mol

Molar mass of NaCl = 58.44 g/mol

Mass = Moles*Molar mass = 0.0075\times 58.44\ g = 0.4383 g

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