What is a personal factor of resilience?

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

4 years ago

A pharmacy intern is asked to prepare 3 L of a 30% w/v solution. T he pharmacy stocks the active ingredient in 8-ounce bottles o

MariettaO [177]

Answer: The number of bottles that will be needed are 6

Explanation:

We are given:

Amount of solution, the intern is asked to prepare = 3 L = 3000 mL (Conversion factor: 1 L = 1000 mL)

Strength of solution needed = 30 % (w/v)

This means that in 100 mL of solution, the solute present is 30 grams

So, in 3000 mL of solution, the solute present will be = $\frac{30}{100}\times 3000=900g$

Active ingredient present in 1 bottle = 8 ounce of 70 % (w/v)

Conversion factor used: 1 ounce = 29.57 mL

So, $8ounce\times \frac{29.57mL}{1ounce}=236.6mL$

Amount of active ingredient present in 1 bottle = $236.6\times \frac{70}{100}=165.6g$

To calculate the number of bottles, we need to divide the total amount of solution needed by the amount of active ingredient present in 1 bottle, we get:

$\text{Number of bottles}=\frac{\text{Amount of solution to prepare}}{\text{Amount of active ingredient in 1 bottle}}$

Putting values in above equation, we get:

$\text{Number of bottles}=\frac{900g}{165.6g}\\\\\text{Number of bottles}=5.43\approx 6$

Hence, the number of bottles that will be needed are 6

6 0

3 years ago

You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,

telo118 [61]

Answer:

$\frac{[F^{-}]}{[HF]}$ is larger

Explanation:

$pK_{a}=-logK_{a}$ , where $K_{a}$ is the acid dissociation constant.

For a monoprotic acid e.g. HA, $K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$ and $\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}$

So, clearly, higher the $K_{a}$ value , lower will the the $pK_{a}$

In this mixture, at equilibrium, $[H^{+}]$ will be constant.

$K_{a}$ of HF is grater than $K_{a}$ of HCN

Hence, $(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})$

So, $\frac{[F^{-}]}{[HF]}$ is larger

8 0

4 years ago

Valence electrons determine an atom’s __________. a. mass c. atomic number b. chemical properties d. period

poizon [28]

Hey there !

Answer B

Chemical properties are characteristic behaviors of a substance when it is found in the presence of another substance. To investigate chemical properties we will have to carry out chemical transformations (performing chemical tests) and it is generally not possible to recover the sample used (because it suffers changes giving rise to other substances).

6 0

4 years ago

Any substance that has_____and takes up space is called matter

Vera_Pavlovna [14]

I believe the answer is Mass, hope this helped

4 0

3 years ago