Where is the most basic building blocks of all organisms

Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)

vichka [17]

Answer:

Kb → 1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m = Kb → 1.56 °C / m

3 0

2 years ago

Reactant → product

KiRa [710]

If we have two gram produce 2 grams were consumed

Due to the mass conversation law which state that the mass in a isolated is neither created nor destroyed by chemical reaction or physical transformation. Therefore the mass of the product must be equal to the mass reactant

5 0

2 years ago

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Use the geometric mean (leg) theorem. What is the

mars1129 [50]

Answer:

2/70

Explanation:

5 0

2 years ago

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The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

2 years ago

¿Cuál es ion monoatomico más estable formado a partir de aluminio?

Naily [24]

Answer:

3+

Explanation:

3 0

2 years ago

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