They were natural philosophers that used experimentation to establish the cause effect relations and explanations, in an epoque when many "theories' were not supported by experiments but by some authority (antique philosophers like Aristoteles or the Catholic Church).

8 0

3 years ago

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Where do the heaviest elements in our solar system come from?

Lady_Fox [76]

The formation of elements heavier than iron and nickel requires the input of energy. Supernova explosions result when the cores of massive stars have exhausted their fuel supplies and burned everything into iron and nickel. The nuclei with mass heavier than nickel are thought to be formed during these explosions.

5 0

3 years ago

Part a cyclohexane has a freezing point of 6.50 ∘c and a kf of 20.0 ∘c/m. What is the freezing point of a solution made by disso

Andrei [34K]

The mathematical expression for the depression in freezing point is given as:

$\Delta T_{f}= k_{f}m$

where, $\Delta T_{f}$ = depression in freezing point

$k_{f}$ = molal depression constant (20.0 $^{o}Ckg/mole$ )

m = molality

Molality is defined as the ratio of number of moles of solute to the kg of the solvent

Number of moles = $\frac{given mass in g}{molar mass}$

given mass of biphenyl solute = 0.925 g

Molar mass of biphenyl = 154.21 g/mol

Put the values,

number of moles of biphenyl = $\frac{0.925 g}{154.21 g/mol}$

= $0.0059 mole$

Molality = $\frac{number of moles of biphenyl solute}{mass of the solvent in kg}$

Mass of cyclohexane solvent = 25.0 g

Convert g into kg: 1 kg = 1000 g

Thus, mass of cyclohexane = $\frac{25.0}{1000}$

= 0.025 kg

Now, put the values in the formula of molality:

Molality = $\frac{0.0059 mole}{0.025 kg}$

= 0.236 mole/kg or 0.236 m.

Calculate the depression in freezing point:

$\Delta T_{f}= 20.0^{o}C kg/mole\times 0.236 mole/kg$

= $4.72 ^{o}C$

Now, $\Delta T_{f} = T_{solvent} - T_{solution}$

$4.72 ^{o}C = 6.50^{o}C - T_{solution}$

$T_{solution} = 6.50^{o}C- 4.72 ^{o}C$

= $1.78^{o}C$

Hence, freezing point of the solution is $1.78^{o}C$ .

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3 years ago

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Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago